 Closed-graph proof for a factored operator

If bounded $S:X\to Y$ is injective and $\mathrm{im}T\subseteq \mathrm{im}S$, then $R=S^{-1}T$ has closed graph: $x_n\to x$ and $R{x}_n\to z$ imply $Sz=Tx=S(Rx)$ and hence $z=Rx$.