Easy Proof Adjoint(Compact)=Compact

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.

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.