It is well known that the graph of a continuous function into a $T_2$ space is closed. Does this result hold if the $T_2$ condition is replaced with $T_1$? I'm guessing the answer is no, but I have been unable to construct a counter-example.
I'm guessing it's false since the proof of the original fact seems to require $Y$ to be Hausdorf quite naturally, you more or less need a pair of open neighborhood of $y$ and $f(x)$ that are disjoint.
So, we're looking for a continuous $f : X \to Y$ where $Y$ is Frechet so that $G = \{(x, f(x)) : x \in X \} \subseteq X \times Y$ is not a closed subset of $X \times Y$ with respect to the product topology.
Certainly, $Y$ must be $T_1$ and not $T_2$, so it's natural to try $Y = \mathbb{N}$ equipped with the cofinite topology, or something similar. If you try and do the simplest possible thing, which is to equip $X$ with the discrete topology, so $f$ is automatically continuous, and pick any function, this won't work. The reason for this is that $$G^C = \bigcup_{x \in X} \left(\{ x\} \times \{ f(x)\}^C \right) $$ means that $G^C$ is in fact open in the product topology, so this doesn't produce a counterexample.
So one must do something more complicated to try and get a counterexample. It seems like some more complicated $X$ could do the trick, but I don't quite know how to work it out.
The cofinite codomain works fine but not with the discrete topology on the domain:
Let $f: (\Bbb R, \tau_e) \to (\Bbb R, \tau_{cf})$ be the identity $f(x)=x$ from the reals in the usual topology to the reals in the cofinite topology. $f$ is continuous as finite sets are closed in $\tau_e$.
If $(x,y) \in \Bbb R^2$ (in the product topology of the domain times the codomain), a basic neighbourhood of it will look be of the form $O:=\langle a,b \rangle \times F^\complement$ where $F$ is a finite subset of the reals and $\langle a,b\rangle$ is an open interval containing $x$ in $\Bbb R$. There is some point $z \in \langle a,b\rangle$ such that $z \notin F$ (as $\langle a,b\rangle \subseteq F$ is impossible) and then $(z,z)$ is a point on the graph of $f$ that intersects $O$.
It follows that the graph of $f$ is dense and far from closed.