How to prove that a bounded linear operator is compact?

The following proof is adapted from Bruce Barnes, Majorization, range inclusion, and factorization for bounded linear operators. One maybe able to simplify the proof somewhat.

Lemma Let $T,S\in B(X,Y)$ and $R(T)\subseteq R(S)$, then $\exists M > 0$ such that for all $\alpha\in Y^*$, $$ \|T^*\alpha\| \leq M\|S^*\alpha\| \tag{1}$$ where $*$ denotes the adjoint operator.

Proof: Let $U$ be the map from $R(S^*) \to X^*$ given by $U(S^*\alpha) = T^*\alpha$. This map is well-defined since the kernel of $S^*$ is contained in that of $T^*$. It suffices to show that $U$ is a bounded linear operator. Suppose not, then there exists a sequence $\alpha_n$ in $Y^*$ such that $S^*\alpha_n$ has norm 1 and $T^*\alpha_n$ diverges. Now take an arbitrary $x\in X$. By assumption there exists $z\in X$ such that $Sz = Tx$. So $$ T^*\alpha_n(x) = \alpha_n(Tx) = \alpha_n(Sz) = S^*\alpha_n(z) $$ and so $$ |T^*\alpha_n(x)| \leq \|z\| < \infty$$ for each $n$. But by the Uniform Boundedness Principle we have that this implies $$ \sup_n \|T^*\alpha_n\| < \infty $$ and we get a contradiction. q.e.d.

Now recall Schauder's Theorem (Dunford and Schwartz, Chapter VI.5, Theorem 2): An operator is compact if and only if its adjoint is compact.

Corollary If in the previous lemma, $S$ is compact, then so is $T$.

Proof: By Schauder's theorem we have that $S^*$ is compact. Since $T^* = US^*$, and $U$ is a bounded linear operator (with bounded linear extension to $R(S^*)$), we have that $T^*$ is a product of a bounded linear operator with a compact operator, and hence is compact. Appealing to Schauder's theorem again we conclude the proof. q.e.d.