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

Since $H$ is a Hilbert space, I will give another proof using this property. 
Assume $T$ is compact and let $(x_n)_{n\in\mathbb{N}}$ be a bounded sequence. As it is bounded, we can construct a weakly convergent subsequence, call it $(x_n)_n$ again in abuse of notation, such that $x_n\to x$ weakly as $n\to\infty$. Now, we want to show that $T^\ast x_n \to T^\ast x$ strongly. Note that
$$\begin{aligned}\|T^\ast (x_n - x) \|^2 &= \langle T^\ast (x_n - x), T^\ast (x_n - x)\rangle=\langle x_n - x , T T^\ast (x_n - x)\rangle \\ &\leq \|x_n-x\| \|T T^\ast (x_n - x)\|\leq C \|T T^\ast (x_n - x)\|,\end{aligned}$$
where we used that a weak convergent sequence is bounded. It suffices to show that $T T^\ast (x_n - x)\to 0 $ strongly as $n\to\infty$. For this, note that $T T^* (x_n - x ) \to 0 $ weakly as $T T^*$ is continuous (in the strong and thus also in the weak topology). Since $T$ is a compact operator and limits are unique we also now that $T T^\ast (x_n - x)\to 0 $ as $n\to\infty$, which concludes the proof.
