Proving that $K:l_2 \to l_2$ is a compact self-adjoint operator I need to prove the following:

Let $a = (a_j)$ be a real sequence from $c_0$ (Banach space of sequences converging to 0) and let X = $l_2$ be a space of complex sequences $(x_j)$ with $\|x\| = (\sum_{j=1}^\infty|x_j|^2)^\frac{1}{2}<\infty$. Define $(Kx)_j = a_jx_j$. Show that $K:X\to X$ is a compact operator.

As I understand, a compact operator is a one whereby for any bounded set $B \in X$, any sequence in $K(B)$ has a convergent subsequence. So let $B$ be bounded and $(y_j) = (x_n)^j$ be a sequence of sequences in $K(B)$. Then we need to find $(y_{j_k})$ a subsequence of $(y_j)$ which converges. But I am not sure how to do this; perhaps we need to extract the compactness from compactness in $\mathbb{C}$? 
As for the self-adjoint part of the question, I could really do with a hint here, too.
Any help is much appreciated!
 A: Start with a sequence of vectors $\{ x_n \}_{n=1}^{\infty}$ for which $\|x_n \| \le M$. Using the Cantor diagonalization process, you can replace $\{ x_n \}$ with a subsequence such that $\lim_k (x_{n_k})_j$ converges for all $j$. So, without loss of generality, assume that $\lim_{n} (x_n)_j$ converges for each $j$. And let $A$ be a constant such that $|a_j| \le A$ for all $j$. Then,
\begin{align}
     \|K x_{n}-Kx_{n+m} \|^2 & = \sum_{j=1}^{\infty}|a_j|^2|(x_n)_j-(x_{n+m})_j|^2 \\
   & \le A\sum_{j=1}^{N}|(x_n)_j-(x_{n+m})_j|^2
     +\sup_{j > N}|a_j|^2\sum_{j=N+1}^{\infty}|(x_n)_j-(x_{n+m})_j|^2 \\
    & \le A\sum_{j=1}^{N}|(x_n)_j-(x_{n+m})_j|^2+4M^2\sup_{j > N}|a_j|^2
\end{align}
Given $\epsilon > 0$, you can choose $N$ large enough that $4M^2\sup_{j > N}|a_j|^2 < \epsilon/2$. Then you can choose $n$ large enough such that, for this fixed $N$, you have $A\sum_{j=1}^{N}|(x_n)_j-(x_{n+m})_j|^2 < \epsilon/2$ because the coordinate sequences converge. It follows that $\{ Kx_n \}_{n=1}^{\infty}$ is a Cauchy sequence and, hence converges.
