An idempotent operator is compact if and only if it is of finite rank

Okay. If you require an answer that uses the sequential definition of a compact operator, here it goes. :)

A finite-rank operator is clearly compact, so one direction is established.

Now, an operator $ T: \mathcal{H} \rightarrow \mathcal{H} $ is compact if and only if it maps every bounded sequence in $ \mathcal{H} $ to a sequence in $ \mathcal{H} $ that contains a convergent subsequence. This is the basic sequential characterization of compact operators.

Let $ T: \mathcal{H} \rightarrow \mathcal{H} $ be an idempotent and compact operator. We want to show that it is a finite-rank operator. By way of contradiction, assume that it is not finite-rank. Let $ R(T) $ denote the infinite-dimensional range space of $ T $. We claim that $ R(T) $ is a closed subspace of $ \mathcal{H} $. To prove this claim, let $ (T(\mathbf{x}_{n}))_{n \in \mathbb{N}} $ be a sequence in $ R(T) $ that converges to some $ \mathbf{y} \in \mathcal{H} $. Then $$ T(\mathbf{y}) = T \left( \lim_{n \rightarrow \infty} T(\mathbf{x}_{n}) \right) = \lim_{n \rightarrow \infty} T(T(\mathbf{x}_{n})) = \lim_{n \rightarrow \infty} T(\mathbf{x}_{n}) = \mathbf{y}. $$ Hence, $ \mathbf{y} \in R(T) $, and as $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ is arbitrary, we are done.

As $ R(T) $ is now seen to be a closed (hence complete) subspace of $ \mathcal{H} $, it follows that $ R(T) $ is a Hilbert subspace of $ \mathcal{H} $. By the infinite-dimensionality of $ R(T) $, pick an orthonormal sequence of vectors $ (\mathbf{e}_{n})_{n \in \mathbb{N}} $ in $ R(T) $ (we can do so by extracting an orthonormal sequence from some infinite orthonormal basis for $ R(T) $). By construction, $ \mathbf{e}_{n} \in R(T) $ for each $ n \in \mathbb{N} $, so we can associate a $ \mathbf{v}_{n} \in \mathcal{H} $ such that $ T(\mathbf{v}_{n}) = \mathbf{e}_{n} $. Then $$ \forall n \in \mathbb{N}: \quad T(\mathbf{e}_{n}) = T(T(\mathbf{v}_{n})) = T(\mathbf{v}_{n}) = \mathbf{e}_{n}. $$ Hence, $ T $ maps the sequence $ (\mathbf{e}_{n})_{n \in \mathbb{N}} $ identically to itself. This sequence is clearly bounded. However, $ (T(\mathbf{e}_{n}))_{n \in \mathbb{N}} = (\mathbf{e}_{n})_{n \in \mathbb{N}} $ does not contain a convergent subsequence! (Simply observe that $ \| \mathbf{e}_{m} - \mathbf{e}_{n} \|_{\mathcal{H}} = \sqrt{2} $ for distinct $ m,n \in \mathbb{N} $.) It follows from the sequential definition of compactness that $ T $ is not compact, which contradicts the fact that we started our argument with $ T $ being compact.

Conclusion $ T $ must be a finite-rank operator.

I noticed that people have been asking questions about the exercises from the chapter on compact operators in Conway's A Course in Functional Analysis. For the benefit of the OP and also of other people who are interested in those exercises, the following link may be useful: A Question on Compact Operators. It contains the full solution to another exercise from the same chapter.

If $P$ is finite-rank, then it is compact.

If $P$ is compact, then its spectrum consists only of eigenvalues, with $0$ as the only possible accumulation point. The equality $P^2=P$ implies that the only eigenvalues are $0$ and $1$. Now take any $x$ in the range of $P$; then $$ Px=x, $$ so $x$ is an eigenvector with eigenvalue $1$. The multiplicity of $1$ has to be finite, as otherwise $P$ would not be compact. Thus the range of $P$ is finite-dimensional.

Let $ T \in B(\mathcal{H}) $ be idempotent.

If $ T $ is of finite rank, then it is clearly compact, so one direction is proven.

Conversely, suppose that $ T $ is compact. By way of contradiction, assume that $ T $ has infinite rank. Then $ T $ acts as the identity operator on its infinite-dimensional range space $ R(T) $. Letting $ \mathbb{B}_{R(T)} $ denote the open unit ball of $ R(T) $, we see that $ T $ maps $ \mathbb{B}_{R(T)} $ identically to itself. However, as $ \mathbb{B}_{R(T)} $ is not relatively compact in $ \mathcal{H} $ (i.e., $ {\text{cl}_{\mathcal{H}}}(\mathbb{B}_{R(T)}) $ is not compact; see below), this contradicts the compactness of $ T $. Therefore, $ T $ must have finite rank.

Notes

As mentioned, $ {\text{cl}_{\mathcal{H}}}(\mathbb{B}_{R(T)}) $ is not compact. If it were compact, then as $ R(T) $ is a closed subspace of $ \mathcal{H} $ (this follows from the idempotence of $ T $) and $ {\text{cl}_{R(T)}}(\mathbb{B}_{R(T)}) = {\text{cl}_{\mathcal{H}}}(\mathbb{B}_{R(T)}) \cap R(T) $, it would follow that $ {\text{cl}_{R(T)}}(\mathbb{B}_{R(T)}) $ is compact. However, this contradicts Riesz's Lemma, which implies that the closed unit ball in any infinite-dimensional normed space is never compact.