Prove: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$ where $\lim_{x \to a} g(x) \neq 0$
If we can show that $\lim_{x \to a} \frac{1}{g(x)} = \frac{1}{\lim_{x \to a} g(x)}$ then we can define $h(x) = \frac{1}{g(x)}$ and appeal to the proof for the product of limits (which we'll assume we've proven already).
Let $L = \lim_{x \to a} g(x)$. Then let
$$\forall \epsilon > 0, \exists \delta_1 > 0 : (0 < |x-a| < \delta_1 \implies |g(x) - L| < \frac{|L|^2 \epsilon}{2})$$
$$\forall \epsilon > 0, \exists \delta_2 > 0 : (0 < |x-a| < \delta_2 \implies |g(x) - L| < \frac{|L|}{2})$$
We wish to show that:
$$\forall \epsilon > 0, \exists \delta > 0 : (0 < |x-a| < \delta \implies \left|\frac{1}{g(x)} - \frac{1}{L}\right| < \epsilon)$$
Next, when $0 < |x-a| < \delta_2$, the triangle inequality lets us see that
$$\left|L\right| = \left|L - g(x) + g(x)\right| \leq \left|L - g(x)\right| + \left|g(x)\right| = \left|g(x) - L\right| + \left|g(x)\right| < \frac{|L|}{2} + \left|g(x)\right|$$
Rerranging, we see that
$$\left|L\right| < \frac{|L|}{2} + \left|g(x)\right| \implies \frac{1}{\left|g(x)\right|} < \frac{2}{\left|L\right|} $$
Then if we set $\delta = \min(\delta_1, \delta_2)$, we conclude
$$\begin{align} \left|\frac{1}{g(x)} - \frac{1}{L}\right| &= \left|\frac{L-g(x)}{Lg(x)}\right| \\ &= \left|\frac{g(x) - L}{Lg(x)}\right| \\ &= \frac{1}{|g(x)|} \cdot \frac{1}{|L|} \cdot \left|g(x) - L\right| \\ &< \frac{2}{\left|L\right|} \cdot \frac{1}{|L|} \cdot \frac{|L|^2 \epsilon}{2} \\ &= \epsilon \end{align}$$
Is my proof correct?
No, it is not correct. At several passages, you have the quantity $|L|-\varepsilon$, which may well be negative. Therefore, it cannot possibly be true that, for instance$$\forall \varepsilon > 0, \exists \delta_1 > 0 : (0 < |x-a| < \delta_1)\implies\bigl|g(x) - L\bigr| < |L| \cdot \varepsilon \cdot\bigl(|L| - \epsilon\bigr).$$
The edited version is correct indeed.