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Check out this paper:

F. B. Jones, Totally discontinuous linear functions whose graphs are connected, November 23, (1940).

Abstract:

Cauchy discovered before 1821 that a function satisfying the equation
$$
f(x)+f(y)=f(x+y)
$$
is either continuous or totally discontinuous. After Hamel showed the existence of a discontinuous function, many mathematicians have concerned themselves with problems arising from the study of such functions. However, the following question seems to have gone unanswered: Since the plane image of such a function (the graph of $y =f(x)$) must either be connected or be
totally disconnected, must the function be continuous if its image is connected? The answer is no.

In particular, Theorem 5 presents a nowhere continuous function $f:Bbb R rightarrow Bbb R$ whose graph is connected.

Whether Conway base 13 function is such an example remains unknown. (at least on MSE; see Is the graph of the Conway base 13 function connected?) It turns out the graph of Conway base 13 function is totally disconnected. See this brilliant answer.

Here is an example for $mathbb R^2 to mathbb R$:

$$f(x,y) = begincases y & textwhen x=0text or x=1 \
x & textwhen xin(0,1)text and y=0 \
1-x &textwhen xin(0,1)text and y=x(1-x) \
x(1-x) & textwhen xnotin0,1text and y/x(1-x) notinmathbb Q \
0 & textotherwise endcases $$

This is easily seen to be everywhere discontinuous. But its graph is path-connected.

A similar but simpler construction, also $mathbb R^2tomathbb R$:

$$ beginalign g(1 + rcostheta, rsintheta) = r & quadtextfor r>0,; thetainmathbb Qcap[0,pi] \
g(rcostheta, rsintheta) =r & quad textfor r>0,; thetainmathbb Qcap[pi,2pi] \
g(x,y) =0 & quadtexteverywhere else endalign $$

There is a simple general strategy for many questions of this type, which is to just try to build a counterexample by transfinite induction. Let's first think about what it means for the graph $G$ of a function $f:mathbbRtomathbbR$ to be disconnected. It means there are open sets $U,VsubsetmathbbR^2$ such that $Ucap G$ and $Vcap G$ are both nonempty and together they form a partition of $G$ (we will say $(U,V)$ separates $G$ in that case). So, to make $G$ connected, we just have to one-by-one rule out every such pair $(U,V)$ from separating it.

So, then, here is the construction. Fix an enumeration $(U_alpha,V_alpha)_alpha<mathfrakc$ of all pairs of open subsets of $mathbbR^2$. By a transfinite recursion of length $mathfrakc$ we define values of a function $f:mathbbRtomathbbR$. At the $alpha$th step, we add a new value of $f$ to prevent $(U_alpha,V_alpha)$ from separating the graph of $f$, if necessary. How do we do that? Well, if possible, we define a new value of $f$ such that the corresponding point in the graph $G$ will either be in $U_alphacap V_alpha$ or not be in $U_alphacup V_alpha$, so $U_alphacap G$ and $V_alphacap G$ will not partition $G$.

If this is not possible, then $U_alpha$ and $V_alpha$ must partition $AtimesmathbbR$ where $AsubseteqmathbbR$ is the set of points where we have not yet defined $f$. Since $mathbbR$ is connected, this means we can partition $A$ into sets $B$ and $C$ (both open in $A$) such that $U_alphacap (AtimesmathbbR)=BtimesmathbbR$ and $V_alphacap (AtimesmathbbR)=CtimesmathbbR$. Now since we have defined fewer than $mathfrakc$ values of $f$ so far in this construction, $|mathbbRsetminus A|<mathfrakc$ and in particular $A$ is dense in $mathbbR$. If $B$ were empty, then $U_alpha$ would have empty interior and thus would be empty, and so $(U_alpha,V_alpha)$ can never separate the graph of $f$. A similar conclusion holds if $C$ is empty, so let us assume both $B$ and $C$ are nonempty. It follows that $overlineB$ and $overlineC$ cannot be disjoint (otherwise they would be a nontrivial partition of $mathbbR$ into closed subsets), so there is a point $xinmathbbRsetminus A$ that is an accumulation point of both $B$ and $C$. Since $xnotin A$, we have already defined $f(x)$. Note now that $(x,f(x))notin U_alpha$, since $U_alpha$ would then contain an open ball around $(x,f(x))$ and thus would intersect $CtimesmathbbR$. Similarly, $(x,f(x))notin V_alpha$. Thus $U_alpha$ and $V_alpha$ already do not contain the entire graph of $f$, and so we do not need to do anything to prevent them from separating it.

At the end of this construction we will have a partial function $mathbbRtomathbbR$ such that by construction, its graph is not separated by any pair of open subsets of $mathbbR^2$, and the same is guaranteed to hold for any extension of our function. Extending to a total function, we get a total function $f:mathbbRtomathbbR$ whose graph is connected. But we can of course arrange in this construction for $f$ to be nowhere continuous; for instance, we could start out by defining $f$ on all the rationals so that the image of every open interval is dense in $mathbbR$. In fact, the construction shows that any partial function $mathbbRtomathbbR$ defined on a set of cardinality less than $mathfrakc$ can be extended to a total function whose graph is connected. (Or even stronger, you could start with any partial function whose domain omits $mathfrakc$ points from every interval, since that is all you need to guarantee that the set $A$ is dense at each step.)

Not an answer

Great question, and I don't have an answer for you, but I've got some small thoughts:

By summing up weighted and displaced copies of $f$, you can get discontinuities at many places. For instance, you could write
$$
F(x) = sum_n in Bbb Z fracf(x-n)1+n^2
$$
That'll have an $f$-like discontinuity at every integer.

Digression
A comment asks whether the graph is still connected. Let me show that it is at $x = 1$ as an example, which should be reasonably compelling for other integer points. (For non-integer points, $F$ is continuous, so we're fine).

Write
beginalign
F(x) &= frac12 f(x-1) + sum_nne 1 in Bbb Z fracf(x-n)1+n^2\
&= frac12 f(x-1) + G_1(x)
endalign
where $G_1$ is a function that's continuous at $x = 1$.

Let's look at the graph of $F$ near $1$, say on the interval $(3/4, 5/4)$. It's exactly
$$
K = (x, frac12 f(x-1) + G_1(x)) mid 3/4 < x < 5/4
$$

Contrast this with the graph of $f$ near $0$, which is
$$
H = (x, f(x)) mid -1/4 < x < 1/4
$$
and which we know (from standard calculus books like Spivak) to be connected.

Now look at the function
$$
S : K to H : (x, y) mapsto (x-1, y - G_1(x))
$$
This is clearly continuous and a bijection (and even extends to a bijection from a (vertical) neighborhood of $K$ to a neighborhood of $H$), so $K$ is also connected.

End of digression

And then for numbers with finite base-2-expansions, you can do the same sort of thing: let
$$
G(x) = sum_k in Bbb Z, k > 0 frac12^k F(2^k x)
$$
and that'll have $f$-like discontinuities at all the points with finite base-2 representations, which is a dense set in $Bbb R$.

But I have a feeling that sliding over to the uncountable-set territory is going to be a lot harder.