It is well known that the closed graph theorem does not directly extend to nonlinear maps: even for functions from $\mathbb{R}$ to $\mathbb{R}$, having closed graph does not imply continuity. But let's consider the following reformulation of the closed graph theorem:
If a linear map between Banach spaces has closed graph, it has a point of continuity.
This does not essentially change the meaning, since a linear map is either everywhere continuous or everywhere discontinuous. But now there is a better chance of nonlinear generalization. Hence, my question:
Suppose that a nonlinear map between separable Banach spaces has closed graph. Does it necessarily have a point of continuity?
Nonseparable counterexample
Since the above seems too good to be true, I tried to find a counterexample. So far, found it only in nonseparable setting.
Let $X$ the space of all bounded functions $x:(0,1]\to\mathbb R$ with the supremum norm. Let $(q_n)_{n=1}^\infty$ be an enumeration of the rationals. Define the function $y=F(x)$ separately on each subinterval $(2^{-n},2^{1-n}]$, $n=1,2,\dots$ as
$$
y(t) = \begin{cases} 1 \quad &\text{if $f(2^nt-1)>q_n$} \\
0 \quad &\text{if $f(2^n t-1)\le q_n$}
\end{cases}
$$
I claim that the map $F:X\to X$ satisfies $\|F(x_1)-F(x_2)\|=1$ whenever $x_1\ne x_2$.
Hence, it is nowhere continuous and its graph is a discrete (in particular closed) set.
Indeed, since $x_1\ne x_2$, there is a point $s\in (0,1]$ and a number $n\in\mathbb N$ such that $q_n$ is strictly between $x_1(s)$ and $x_2(s)$; according to the definition of $F$ this implies that the functions $F(x_1)$ and $F(x_2)$ take on different values at the point $t=2^{-n}(s+1)$.
Finite-dimensional case
In finite dimensions, a map with a closed graph is continuous outside of a closed set with empty interior. The proof is not very interesting, so I link to it instead of adding it to the post.