Prove that if $\{s_n\}\in \ell ^2$, then $\lim_{n\to\infty}s_n=0$

Prove that if $\{s_n\}\in \ell ^2$, then $\lim_{n\to\infty}s_n=0$

I'm having a little trouble coming up with the wording for this proof. Here's what I've done so far:

PROOF: Given that $\{s_n\}\in\ell^2$, we know that $$\sum_{n=1}^\infty s_n^2 < \infty \implies \lim_{n\to\infty}s_n^2 = 0.$$ In other words, for any $\epsilon > 0$ there exists $N\in\mathbb N$ such that for all $n\ge N$, $$|s^2_n - 0| = |s_n^2| = s_n^2 < \epsilon.$$ So, observe that $$|s_n - 0| = |s_n| = \sqrt{s_n^2} < \underset{***}{\sqrt\epsilon} \le \epsilon.$$ Therefore, $$\lim_{n\to\infty}s_n = 0.$$

Is the step $***$ valid? I think so, but would the same $N$ from before work here to give us this inequality using the square root?

(***) is not valid since $\sqrt{\epsilon}$ is not necessarily smaller than $\epsilon$ (e.g., consider $\sqrt{0.5}\approx 0.7$).
Hint: Suppose $s_n^2 \rightarrow 0$. Then, for each $\epsilon>0$, $s_n^2<\epsilon^2$ for sufficiently large $n$.