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.
It seems that the statement as is, is not true in general. Additional uniform conditions on the kernel $K$ are required for $T_K$ to be compact.
Consider the kernel
$$ K(t,s)=\frac{1}{\sqrt{st}}\mathbb{1}(t<s)\mathbb{1}(0\leq s,t\leq1)$$
It satisfies the conditions of the problem for $\int^1_0|K(t,s)|\,dt\equiv2$. We have that
$$\int^1_0|K(t+h,s)-K(t,s)|\,dt = 2\mathbb{1}(0<s\leq h) + \frac{2\sqrt{h}}{\sqrt{s}}\mathbb{1}(h<s\leq1)$$
Consequently
\begin{align}
\int^1_0|T_Kx(t+h)-T_Kx(t)|\,dt &\leq \int^1_0|x(s)|\int^1_0 |K(t+h,s)-K(t,s)|\,dt\,ds \\
&=2\int^h_0|x(s)|\,ds +2\sqrt{h}\int^1_h\frac{|x(s)|}{\sqrt{s}}\,ds
\end{align}
From this estimate, one sees that the term $\int^h_0|x(s)|\,ds$ cannot be controlled uniformly on $x\in M$ (unless M is finite or uniformly
integrable).
The problem, in my view, stems from the lack of uniformity with which one can approximate a kernel $K$ of the type described in the problem by nice functions (continuous functions for example). That is, a given $K$ that satisfies the condition $\|\int^1_0K(t,\cdot)\,dt\|_\infty<\infty$ may not by approximated nicely by continuous functions so that
\begin{align}
\int^1_0|x(s)|\int^1_0 |K(t+h,s)-K(t,s)|\,dt\,ds &\leq \int^1_0|x(s)|\int^1_0 |K(t+h,s)-H(t+h,s)|\,dt\,ds + \\
&\quad \int^1_0|x(s)|\int^1_0|H(t+h,s)-H(t,s)|\,dt\,ds +\\
&\quad\int^1_0|x(s)|\int^1_0 |H(t,s)-K(t,s)|\,dt\,ds\\
&\leq \int^1_0|x(s)|\int^1_0|H(t+h,s)-H(t,s)|\,dt\,ds + \\
&\quad 2\int^1_0|x(s)|\int^1_0 |H(t,s)-K(t,s)|\,dt\,ds.
\end{align}
with $H\in C([0,1]\times[0,1])$, becomes uniformly small in $M$.
Here is another example which shows that even weak compactness in $L_1[0,1]$ may not hold either. Define the kernel
$$G(t,s)=\frac{1}{\sqrt{s}}\mathbb{1}(t\leq\sqrt{s})\mathbb{1}(0<s,t\leq1)$$
Then $\int^1_0|G(t,s)\,dt=1$, and $T_Gx(t):=\int^1_0x(s)G(t,s)\,ds=\int^1_{t^2}\frac{x(s)}{\sqrt{s}}\,ds$
The collection of maps $g_s:t\rightarrow G(t,s)$ is not uniformly integrable and in turn, this implies that for a general bounded set $M\subset L_1[0,1]$, $T(M)$ is not uniformly integrable.
Best Answer
a) Let $$f_n(x)=\left\{\begin{array}{cl} 2^nx, & x\in[0,2^{-n}]\\1\ ,& x\in[1-2^{-n},1]\end{array}\right..$$ Then $f_n\in C([0,1])$ and $\|f_n\|_{C([0,1])}=1$. Denote $g_n=Hf_n$. Then
$$g_n(x)=\left\{\begin{array}{cl} 2^{n-1}x, & x\in[0,2^{-n}]\\1-2^{-n-1}x^{-1},& x\in[1-2^{-n},1]\end{array}\right..$$ Note that if $m<n$, then $g_m(2^{-n})=2^{m-n-1}\le\frac{1}{4}$, but $g_m(2^{-m})=\frac{1}{2}$, which implies that $(g_n)$ has no Cauchy subsequence.
b) $H:C([0,1])\to L^2([0,1])$ is compact. Let $(f_n)$ be a sequence in the closed unit ball of $C([0,1])$ and denote $g_n=Hf_n$. It suffices to show that $(g_n)$ has convergent subsequence in $L^2([0,1])$.
Denote $F_n(x)=\int_0^xf_n(t)dt$. Since $(f_n)$ is uniformly bounded, $(F_n)$ is uniformly bounded and equicontinous. Then by Arzelà –Ascoli theorem, $F_n$ has some subsequence convergent in $C([0,1])$. Without loss of generality, let us assume that $\lim_{n\to\infty}F_n=F$ in $C([0,1])$, and we only need to show that $g_n$ converges to $g(x)=F(x)/x$ in $L^2([0,1])$.
Note that $|F_n(x)|\le x$ for every $x\in[0,1]$, so $|F(x)|\le x$, $|g_n(x)|\le 1$ and $|g(x)|\le 1$. Then for every $\delta>0$, $$\int_0^1|g_n(x)-g(x)|^2dx= \int_0^\delta|g_n(x)-g(x)|^2dx+\int_\delta^1|g_n(x)-g(x)|^2dx$$ $$\le 4\delta+\delta^{-2}\int_\delta^1|F_n(x)-F(x)|^2dx.$$ First letting $n\to\infty$ and then letting $\delta\to 0$, the conclusion follows.