Easy Proof Adjoint(Compact)=Compact I am looking for an easy proof that the adjoint of a compact operator on a Hilbert space is again compact. This makes the big characterization theorem for compact operators (i.e. compact iff image of unit ball is relatively compact iff image of unit ball is compact iff norm limit of finite rank operators) much easier to prove, provided that you have already developed spectral theory for C*-algebras. 
By the way, I'm using the definition that an operator $T\colon H \to H$ is compact if and only if given any [bounded] sequence of vectors $(x_n)$, the image sequence $(Tx_n)$ has a convergent subsequence. 
edited for bounded

What you're asking about is called Schauder's theorem.

An operator $T: X \to Y$ is compact if and only if $T^{\ast}: Y^{\ast} \to X^{\ast}$ is compact.

I'm using the following definition of compactness: An operator $T: X \to Y$ between Banach spaces is compact if and only if every sequence $(x_{n}) \subset B_{X}$ in the unit ball of $X$ has a subsequence $(x_{n_{j}})$ such that $(Tx_{n_j})$ converges. This implies that $K = \overline{T(B_{X})} \subset Y$ is compact, as it is sequentially compact and metric. Now let $(\phi_{n}) \subset B_{Y^{\ast}}$ be any sequence and we want to show that $(T^{\ast}\phi_{n})$ has a convergent subsequence. Observe that the sequence $f_{n} = \phi_{n}|_{K}$  in $C(K)$ is bounded and equicontinuous, so by the theorem of Arzelà-Ascoli, the sequence $(f_{n})$ has a convergent subsequence $(f_{n_{j}})$ in $C(K)$. Now observe 
$$\|T^{\ast}\phi_{n_i} - T^{\ast}\phi_{n_{j}}\| = \sup_{x \in B_{X}} \|\phi_{n_i}(Tx) - \phi_{n_j}(Tx)\| = \sup_{k \in K} |f_{n_i}(k) - f_{n_j}(k)|$$
where the last equality follows from the fact that $T(B_{X})$ is dense in $K$. But this means that $(T^{\ast}\phi_{n_j})$ is a Cauchy sequence in $X^{\ast}$, hence it converges.
I leave the other implication as well as the translation to the Hilbert adjoint to you as an exercise.
