Let $H$ be a complex Hilbert space with linear subspaces $U,V$. Then a (not necessarily bounded) linear operator $T:U\to V$ is said to be closed if $$\text{Graph}(T)\equiv\lbrace(u,Tu):u\in U\rbrace$$ is closed in $H\oplus H$.
Now let $U,V,W$ be linear subspaces of $H$ and $S:U\to V$, $T:V\to W$ be closed (not necessarily bounded) linear operators. Clearly $T\circ S:U\to W$ is a linear operator. We have that $\text{Graph}(S)=\lbrace(u,Su):u\in U\rbrace$ and $\text{Graph}(T)\equiv\lbrace(v,Tv):v\in V\rbrace$ are closed in $H\oplus H$ and need to show that $\text{Graph}(TS)\equiv\lbrace(u,TSu):u\in U\rbrace$ is closed in $H\oplus H$.
So if $(u_n,TSu_n)\in\text{Graph}(TS)$ with $(u_n,TSu_n)\to(x,y)\in H\oplus H$, then we need to show that $(x,y)\in\text{Graph}(TS)$ i.e. that $y=TSx$. We find that $u_n\to x$ and $TSu_n\to y$. Clearly $y=TSx$ if $S$ and $T$ are bounded, but I am looking at unbounded operators here. How can I use the fact that $S$ and $T$ are closed to prove that $TS$ is closed when $S$ and $T$ are unbounded/not continuous?
First, note that $$X=\{(u,Su,TSu):u\in U\}\subseteq H^3$$ is both closed and convex. Thus it is weakly closed, and (since $H$ is reflexive) any bounded subset of $X$ is weakly compact.
Second, let $\pi_{1,3}$ denote the projection $H\oplus H\oplus H\to H\oplus H$ that omits the middle direct summand. Clearly, $\pi_{1,3}X=\mathrm{Graph}(TS)$. Since continuous maps send (weak) compacta to compacta, any bounded subset of $\mathrm{Graph}(TS)$ is (weakly) compact. In particular, any such set $Y$ is weakly (and so strongly) closed.
Finally, suppose $\{(u_n,TSu_n)\}_n\to(v,w)$. Then $\{(u_n,TSu_n)\}_n$ is bounded; choose bounded compact $Y\supseteq\{(u_n,TSu_n):n\}$. Since $Y$ is closed, $(v,w)\in Y\subseteq\mathrm{Graph}(TS)$.