I'd like someone to verify, if my below proof is technically correct and sufficiently rigorous. Are the lower bounds for $|g_n(x)|$ and $|g(x)|$ deduced correctly?
[Abbott 6.2.8] Let $(g_n)$ be a sequence of continuous functions that converges uniformly to $g$ on a compact set $K$. If $g(x) \neq 0$ on $K$, show that $(1/g_n)$ converges uniformly on $K$ to $1/g$.
Proof. (My Attempt)
Let's explore the expression
\begin{align*}\left| \frac{1}{g_n(x)} - \frac{1}{g(x)}\right| = \frac{|g_n(x) - g(x)|}{|g_n(x)||g(x)|}\end{align*}
By the continuous limit theorem, since $(g_n)$ converges uniformly to $g$ on $K$ and each $g_n$ is continuous for all $n\in \mathbf{N}$, it follows that $g$ is continuous over the set $K$. As $g$ is continuous over a compact set $K$, by the extreme value theorem, $m=\inf g(K)$ exists, $m \in g(K)$ and further $m \neq 0$. We have, $|g(x)| \geq |m|$ for all $x\in K$.
Consider the particular value $\epsilon_0 = \frac{|m|}{2}$. Since, $(g_n)$ converges uniformly to $g$, there exists $N_1(\epsilon)$, such that
$$||g_n(x)| - |g(x)|| \leq |g_n(x) - g(x)| < \frac{|m|}{2}$$
for all $n \geq N_1$ and for all $x \in K$. So,
\begin{align*}|g_n(x)| &> |g(x)| - \frac{|m|}{2}\\&> \left(|m| - \frac{|m|}{2}\right) = \frac{|m|}{2}\end{align*} Thus,
\begin{align*}\left|\frac{1}{g_n(x)} - \frac{1}{g(x)}\right| &< \frac{|g_n(x) - g(x)|}{\frac{m^2}{2}}\end{align*}
for all $x \in K$ and $n \geq N_1$.
Now, select $N_2(\epsilon)$, so that
$$|g_n(x) - g(x)| < \frac{m^2}{2} \cdot \epsilon$$
for all $n \geq N_2$ and $x \in K$.
Let $N = max \{N_1,N_2\}$. We have,
$$\left|\frac{1}{g_n(x)} - \frac{1}{g(x)}\right| < \epsilon$$
for all $n \geq N$ and $x \in K$. Consequently, the sequence $(g_n)$ converges uniformly to $g$.