$(\Rightarrow)$: Assume $T$ is compact on $\mathcal{H}$ and $\{e_{n}\}$ is an orthonormal basis of $\mathcal{H}$. Define:

It is clear that the $P_{n}$ have finite rank and that we have $\|P_{n}f\|\leq\|f\|$ for all $n\in\mathbb{N}$, $f\in\mathcal{H}$.

Let $\mathcal{B}$ be the unit ball in $\mathcal{H}$. We have that $P_{n}\to I$ pointwise. Since the $P_{n}$ are contractive they are equicontinuous, hence $P_{n}$ converges uniformly to $I$ on compact sets, and in particular on $\overline{T(\mathcal{B})}$, which is compact by assumption. Therefore $P_{n}T\to T$ uniformly on $\mathcal{B}$, hence $\|P_{n}T-T\|\to 0$. Since $P_{n}T$ is bounded and of finite rank the first direction follows.

$(\Leftarrow)$: Now let $\{F_{n}\}$ be a sequence of bounded operators of finite rank with $\|T-F_{n}\|\to 0$. We have to show that $T(\mathcal{B})$ is relatively compact in $\mathcal{H}$. This is equivalent to $T(\mathcal{B})$ being totally bounded in $\mathcal{H}$. So we are left to show that for all $\epsilon>0$ there is an $\epsilon$-net $x_{1},\cdots,x_{n}\in\mathcal{H}$ so that:

So choose $\epsilon>0$ and $n\in\mathbb{N}$ fixed so that:

Choose $x_{1},\cdots,x_{m}\in\mathcal{H}$ with:

Hence (by the triangle inequality):

and we are done. ∎