If $T:X\to Y$ is a compact linear operator on Banach spaces $X$ and $Y$and $S:Y\to Z$ is a bounded linear operator where $Z$ is a Banach space. Prove $ST$ is compact.
Here's my proof, let $x_n \in X$ be a sequence such that $\| x_n\| \leq 1$ , then since $T$ is compact then $Tx_n$ has a convergent subsequence say $Tx_{n_k} \to y$, Then since $S$ is bounded we get $STx_{n_k} \to Sy$ which shows that $ST$ has a convergent subsequence proving that $ST$ is compact.
Is my proof correct?