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?
Alternate proof: Let $A$ be any bounded subset of $X$. Then $Cl_Y(T[A])$ is compact in $Y$. Now $S$ is continuous because $S$ is bounded, and the continuous image of a compact set is compact,
so $S[Cl_Y(T[A])]$ is compact in $Z.$
In a normed linear space (& in any Hausdorff space), compact subsets are closed. So $$S[Cl_Y(T[A])]=Cl_Z(S[Cl_Y(T[A])]).$$ We now have $$Cl_Z((ST)[A])=Cl_Z(S[T[A]])\subseteq$$ $$\subseteq Cl_Z(S[Cl_Y(T[A])])=$$ $$=S[Cl_Y(T[A])].$$ So $Cl_Z((ST)[A])$ is a closed subset of the compact set $S[Cl_Y(T[A])],$ so $Cl_Z((ST)[A])$ is compact.
The proposer's proof is also correct.