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.
To directly show that $V$ maps bounded sets to precompact sets, the only idea I have would be to replicate most of the proof of Ascoli's theorem.
We can show it semi-directly, though, by exposing $V$ as the norm-limit of operators with finite-dimensional range.
For $1 \leqslant k < n$, let
$$\lambda_{n,k}(f) = \int_0^{k/n} f(t)\,dt,$$
and
$$\chi_{n,k}(x) = \begin{cases} 1 &, \frac{k}{n} \leqslant x < \frac{k+1}{n}\\ 0 &, \text{ otherwise}. \end{cases}$$
Define
$$V_n(f) = \sum_{k=1}^{n-1} \lambda_{n,k}(f)\cdot \chi_{n,k}$$
for $n \geqslant 1$. Then $V_n$ is a continuous operator with finite-dimensional range. For $0 \leqslant k < n$ and $\frac{k}{n} \leqslant x < \frac{k+1}{n}$, we have
$$\lvert V(f)(x) - V_n(f)(x)\rvert = \left\lvert\int_{k/n}^{x} f(t)\,dt \right\rvert
\leqslant \sqrt{x-\frac{k}{n}}\cdot \lVert f\rVert_{L^2} \leqslant \frac{1}{\sqrt{n}}\lVert f\rVert_{L^2},$$
and hence $\lVert V - V_n\rVert \leqslant \frac{1}{\sqrt{n}}$. Thus $V$ is the norm-limit of operators with finite-dimensional rank, therefore compact.
Best Answer
This is how I would do it:
First assume that $\{a_n\}$ does not converge to zero. This means that there exists $\varepsilon>0$ and a subsequence $\{a_{n_k}\}_k$ with $|a_{n_k}|\geqslant\varepsilon$. Now consider the sequence of vectors $\{e_k\}$, where $e_k$ has a 1 in the $n_k$ position, and zero elsewhere. Then $Te_k$ is the sequence with $a_{n_k}$ in the $n_k$-entry and zeroes elswhere. So $\|Te_k-Te_j\|_2\geqslant\sqrt2\varepsilon$; considering the balls of radius $\varepsilon/2$ centered on the $Te_k$, we produce an infinite number of disjoint balls in $\overline{T(B_X)}$, which shows that $\overline{T(B_X)}$ is not compact, i.e. $T$ is not compact. This proves that if $T$ is compact, then the sequence goes to zero.
Now assume that $\lim a_n=0$. Let $y_1,y_2,\ldots$ be a sequence in $\overline{T(B_X)}$. Fix $\varepsilon>0$. Then we can get a sequence $x_1,x_2,\dots$ in $B_X$ with $\|y_j-Tx_j\|_2<2^{-j}\varepsilon$ for all $j$. Fix $n_0$ such that $|a_n|<\sqrt{\varepsilon/8}$ when $n\geqslant n_0$. Now, for each $k=1,\ldots,n_0$, consider the sequence of $k^{\rm th}$ entries of the sequence $\{x_j\}_j$. As this is a finite number of sequences in the unit ball of $\mathbb{C}$, there is a subsequence $\{x_{j_h}\}_h$ such that its first $n_0$ entries converge. So we can find $h$ such that, for $\ell=1,\ldots,n_0$, $$ |x_{j_{h+m}}(\ell)-x_{j_h}(\ell)|<\frac{\sqrt\varepsilon}{2^{(\ell+1)/2}K^{1/2}}\ \ \ \text{ for all }m $$ (i.e. $\{x_{j_h}\}$ is Cauchy in its first $n_0$ coordinates). Then $$ \|Tx_{j_{h+m}}-Tx_{j_h}\|_2^2=\sum_{\ell=1}^{n_0}|a_\ell(x_{j_{h+m}}(\ell)-x_{j_h}(\ell))|^2 +\sum_{\ell=n_0+1}^\infty|a_\ell(x_{j_{h+m}}(\ell)-x_{j_h}(\ell))|^2 \\ \leqslant\frac{\varepsilon}2+\frac{\varepsilon}8\,\|x_{j_{h+1}}-x_{j_h}\|_2^2 \leqslant\frac{\varepsilon}2+\frac{\varepsilon}8\,2^2=\varepsilon. $$ We have shown that $\{Tx_{j_h}\}_h$ is Cauchy, and so it is convergent in $\overline{T(B_X)}$. The sequence $\{y_{j_h}\}_h$ gets arbitrarily close to this sequence, so it is also convergent. So $T$ is compact.