Is there $f:\mathbb{R}\to\mathbb{R}$ such that the graph is connected but for any two points in the graph, there isn´t any path in the graph between them?
I´m not sure what the answer is going to be. While trying to prove that there is such a function, it only occurred to me using functions such that the image of any open set is all the real line, but that seems not to be enough to ensure that the graph is connected.
Edit: The function exists, as shown in the comments of Brandon du Preez and HallaSurvivor. In fact, there are solutions $f$ with $f(x)+f(y)=f(x+y)\forall x,y$.
As explained in the comments by Brandon du Preez and HallaSurvivor, such functions exist. Indeed, the following paper:
F. B. Jones, Totally discontinuous linear functions whose graphs are connected, November 23, (1940).
proves the existence of a function $f:\mathbb{R}\to\mathbb{R}$ which has a connected graph but it is totally discontinuous.
If you pick distinct points in the graphs of $f$, $(x,f(x))$ and $(y,f(y))$, with $x>y$, there can be no path between them in the graph. This is because if there was a path between them, there would be an arc $c:[0,1]\to\mathbb{R}^2$ from $(x,f(x))$ to $(y,f(y))$, due to Hausdorff, pathwise connected spaces being arcwise connected (this is Corollary. 31.6 in $\textit{General topology}$, by Willard).
Now calling $\pi_1,\pi_2:\mathbb{R}^2\to\mathbb{R}$ the projections onto the factors, the function $g=\pi_1\circ c:[0,1]\to[x,y]$ would be injective and continuous, and thus an homeomorphism. So $f|_{[x,y]}=\pi_2\circ c\circ g^{-1}:[x,y]\to\mathbb{R}$ would be continuous, contradicting the fact that $f$ is totally discontinuous.