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.
Your intuition about the restriction of a compact operator to a subspace looks correct to me. A linear operator $T : X \to Y$ is compact if, for every bounded sequence $\{x_n\}$ in $X$, the sequence $\{Tx_n\}$ has a subsequence which is Cauchy in $Y$. So, if you have a subspace $Z$ of $X$, any bounded subsequence in $Z$ is also a bounded subsequence in $X$, so its image under $T$ will have a Cauchy subsequence in $Y$ too.
For your second question, I don't think the range of a compact operator is always closed. Consider the operator $T : \ell^2 \to \ell^2$,
$ (Tx)_n = e^{-|n|}x_n$.
(Pretty sure that's a compact operator.) For all $y$ in the range of $T$, $y_n$ decays exponentially fast as $n \to \pm\infty$. Now consider the sequence $z_n = (1+n^2)^{-1}$, which is in $\ell^2$; since it doesn't decay exponentially fast, it can't be in the range of $T$. On the other hand, there is a sequence of points in the range of $T$ which converge to $z$. If you're stumped, I can write down what the sequence is.
Someone who's less rusty on functional analysis can probably give you a deeper insight here. You should note that an operator have a closed range and being a closed operator are two different things. For example, it's shown here that an operator is closed if its domain is complete with respect to the graph norm
$\|x\|_T = \sqrt{\|x\|^2+\|Tx\|^2}$.
You can verify that, by this criterion, the counterexample I gave you before is a closed operator, even though its range isn't closed.
Best Answer
Suppose that $g$ is not zero a.e. and let $\epsilon>0$ be such that $E=\{x:|g(x)|>\epsilon\}$ has nonzero measure. Consider the orthogonal projection $p=T_{\chi_{E}}$. Assume that $T_{g}$ is a compact operator. The operator $T_{g}$ naturally induces a continuous linear operator of $L^2(\mathbb{R})$ into the Hilbert subspace $pL^{2}(\mathbb{R})=\{\xi \in L^{2}(\mathbb{R}):p\xi=\xi\}$, which is a Banach space. This map is onto, since $\xi \in pL^{2}(\mathbb{R})$ is mapped to by the $L^{2}$ function equal to $\xi(x)/g(x)$ on $E$ and zero elsewhere. By the open mapping theorem we have that this induced map is open, and therefore the image of the unit ball of $L^{2}(\mathbb{R})$ under this induced map contains an open ball $B$, and therefore the closure contains a closed ball $\overline{B}$. This closed ball is not compact, since $pL^2(\mathbb{R})$ is infinite-dimensional. But the image of the unit ball of $L^2(\mathbb{R})$ under a compact operator must have compact closure, and closed subsets of compact sets must be compact, so this is a contradiction.