Statement: Let $T \in \mathscr{B}(\mathscr{H})$, where $T$ is a compact operator. Let $M$ be a closed invariant subspace of $T$. Show that the restriction of $T$ to $M$ is compact.
Attempted Proof: Let $\{m_n\}_{n=1}^{\infty}$ be a bounded sequence in $M$. Since $T$ is compact it contains a convergent subsequence $T(m_{n_k}) \rightarrow \mu$. Now since $M$ is closed and invariant under $T$ it contains all of its limit points, so $\mu \in M$. Hence we are compact.
Q.E.D.
The proof is good. However, the phrasing can be improved a little. For example, instead of "Since $T$ is compact it contains a convergent..." we have to replace "it" by "the sequence $\{T(m_n)\}$".
At the end, instead of saying "we are compact", maybe we could replace "we" by "the restriction of $T$ to $M$".