Output of a stable LTI system (discrete)
So I'm answering my question here; turned out that I have to solve (2) then solve (1).

To prove $y[n]$ has finite energy, we write

$$\sum_{n=-\infty}^{\infty}y^2[n] = \sum_{n=-\infty}^{\infty}(\sum_{k=-\infty}^{\infty}h[k]x[n-k])^2 = \sum_{n=-\infty}^{\infty}(\sum_{k=-\infty}^{\infty}h[k]x[n-k])(\sum_{l=-\infty}^{\infty}h[l]x[n-l])$$
$$ = \sum_{k=-\infty}^{\infty}\sum_{l=-\infty}^{\infty}h[k]h[l]\sum_{n=-\infty}^{\infty}x[n-k]x[n-l]$$
Now we know that $x[n]$ has finite energy, so
$$ \sum_{n=-\infty}^{\infty}x[n-k]x[n-l] \leq \sum_{n=-\infty}^{\infty}x^2[n] = E$$
(a generalization of $a^2 + b^2 \geq 2ab$)
Then
$$ \sum_{n=-\infty}^{\infty}y^2[n] \leq E\sum_{k=-\infty}^{\infty}|h[k]|\sum_{l=-\infty}^{\infty}|h[l]| $$
For a BIBO system, $$\sum_{l=-\infty}^{\infty}|h[l]| \leq M \leq \infty$$
So we approve $$\sum_{n=-\infty}^{\infty}y^2[n] \leq M^2E \leq \infty$$

To prove $x[n]$ tends to a constant:

$$y[n] = a\sum_{k=-\infty}^{\infty}h[k]u[n-k] + \sum_{k=-\infty}^{\infty}h[k]x_0[n-k] = a\sum_{k=-\infty}^{\infty}h[k]+y_0[n]$$
Using the conclusion of 1, we know that $$\sum_{n=-\infty}^{\infty}x^2[n] &lt; \infty \rightarrow \sum_{n=-\infty}^{\infty}y^2[n] &lt; \infty$$
which implies that $$\lim_{n \rightarrow \infty} |y_0[n]| = 0.$$
So
$$ \lim_{n \rightarrow \infty} y[n] = a \times \lim_{n \rightarrow \infty} \sum_{k=0}^{\infty}h[k] = ac$$
where $c$ is a constant (because of BIBO).