The limit of a product of functions equals the product of the limits: Is this proof rigorous?

Yes, the second proof is rigorous enough. But for clarity, it should add at the end:

Hence for any positive $\varepsilon'$, there exists a positive $\varepsilon < \varepsilon'$ that satisfies conditions (1) and (2).

This is important because you need $\varepsilon' \rightarrow 0 \Rightarrow\varepsilon \rightarrow 0$

You can simplify it if you assume that $\varepsilon < 1$.

Then, from $|f(x)g(x)-FG| < \varepsilon^2 + \varepsilon(|F| + |G|) $ (you wrote $|L|$ instead of $|F|$) you get $|f(x)g(x)-FG| < \varepsilon(1+|F| + |G|) $ so you can just choose $\varepsilon \lt \dfrac{\min(1, \varepsilon')}{1+|F| + |G|} $.