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After noticing that function $f: mathbb Rrightarrow mathbb R $ $$ f(x) = left{beginarrayl sinfrac1x & textfor xneq 0 \ 0 &textfor x=0 endarrayright. $$
has a graph that is a connected set, despite the function not being continuous at $x=0$, I started wondering, doest there exist a function $f: Xrightarrow Y$ that is nowhere continuous, but still has a connected graph?

I would like to consider three cases

• 
  $X$ and $Y$ being general topological spaces


• 
  $X$ and $Y$ being Hausdorff spaces


• ADDED: $X=Y=mathbb R$
  

But if you have answer for other, more specific cases, they may be interesting too.

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 $mathbb R^2tomathbb R$:

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

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.