Why is this proof incorrect? (limit product is product of the limits)

I always wonder why people insist in cutting epsilons in pieces.

Just remember that eventually proving $\forall n>N, |u_n-\ell|<K\varepsilon$ is sufficient as long as $K$ is constant.

Trying to cut epsilons is maybe aesthetically nice but, I think it hurts the understanding at basic level.

Instead with the straightforward proof you get to : $\begin{cases} \forall n>N_1, |s_n-L_1|<\varepsilon\\ \forall n>N_2, |t_n-L_2|<\varepsilon\\ \end{cases}$

So for $N>\max(N_1,N_2)$

we have $|s_nt_n-L_1L_2|\le|s_n||t_n-L_2|+|L_2||s_n-L_1|\le\left(|s_n|+|L_2|\right)\varepsilon$

Now you see that you do not have a constant before $\varepsilon$, and get to think about why $(s_n)_n$ should be bounded.

And indeed, any convergent sequence is bounded, thus $|s_n|<M$ independently of $n$.

You arrive to $|s_nt_n-L_1L_2|<\underbrace{(M+|L_2|)}_{\text{a constant }K}\varepsilon$

And you should be happy with that, it is not mandatory to get to a bare $\varepsilon$ in the end, $0.0003\,\varepsilon,\ 210734\,\varepsilon,\ 10^{513}\,\varepsilon$ or $K\varepsilon$ are all the same.

Example:

Take the sequence $a_n=1/n$. It converges to zero, but the statement "There exists some $N\in \Bbb N$ such that $n>N$ implies $|a_n|<\epsilon/n^2$" is false. So, indeed, $\epsilon$ can't depend on $n$.

Fixing the proof:

To fix your proof, take an upper bound $M$ of $|s_n|$, which must exist because $s_n$ converges, and write $M$ instead of $|s_n|$ in those denominators.

Your proof is almost correct. To avoid $n$- dependence and zero denominators just replace the bound $\dfrac{\epsilon}{|s_n|+|L_2|}$ with $\dfrac{\epsilon}{M+|L_2|}$ where $M$ is a real positive number such that $|s_n|<M$ for all $n$. Since $(s_n)_n$ is convergent it is bounded and therefore the number $M$ exists.