Here is Prob. 10 (a), Sec. 26, in the book Topology by James R. Munkres, 2nd edition:
Prove the following partial converse to the uniform limit theorem:
Theorem. Let $f_n \colon X \to \mathbb{R}$ be a sequence of continuous functions, with $f_n(x) \to f(x)$ for each $x \in X$. If $f$ is continuous, and if the sequence $f_n$ is monotone increasing, and if $X$ is compact, then the convergence is uniform. [We say that $f_n$ is monotone increasing if $f_n(x) \leq f_{n+1}(x)$ for all $n$ and $x$.]
Here is another post of mine here on Math Stack Exchange on this very problem, where I've also given a proof of the analogue of this very result. In what follows, I'll try to give a different proof of this result.
My Attempt:
Let $\varepsilon > 0$ be given.
For each $x \in X$, as the sequence $f_n(x)$ is a monotonically increasing, convergent sequence of real numbers with limit $f(x)$, so we must have $$ f(x) = \sup \left\{ \ f_n(x) \ \colon \ n \in \mathbb{N} \ \right\}. \tag{0} $$ Moreover, the set $$ \big( \ f(x) - \varepsilon, \, f(x) + \varepsilon \ \big) = \{ \ y \in \mathbb{R} \ \colon \ f(x) - \varepsilon < y < f(x) + \varepsilon \ \} $$ is an open subset of $\mathbb{R}$, and as $f$ is continuous, so the inverse image $$U_x \colon= f^{-1} \left( \ \big( \ f(x) - \varepsilon, \, f(x) + \varepsilon \ \big) \ \right) = \{ \ u \in X \ \colon \ f(x) - \varepsilon < f(u) < f(x) + \varepsilon \ \} \tag{1} $$ is an open set in $X$ and $x \in f^{-1} \left( \ \big( \ f(x) - \varepsilon, \, f(x) + \varepsilon \ \big) \ \right)$, because $$ f(x) - \varepsilon < f(x) < f(x) + \varepsilon \tag{2} $$ holds. Furthermore, as $$ \lim_{n \to \infty} f_n(x) = f(x), $$ so there exists a natural number $N_x$ such that $$ \left\lvert f_n(x) - f(x) \right\rvert < \varepsilon \tag{3} $$ for all natural numbers $n > N_x$.
The collection $\left\{ \ U_x \ \colon \ x \in X \right\}$ in (1) above is an open covering of $X$, and as $X$ is compact, so there are finitely many sets $U_{x_1}, \ldots, U_{x_K}$, say, such that $$ X = \bigcup_{i=1}^K U_{x_i}. \tag{4} $$
Now let us put $$ N \colon= \max \left\{ \ N_{x_1}, \ldots, N_{x_K} \ \right\}. \tag{5} $$ Now for any natural number $n > N$, we have $n > N_i$ for each $i = 1, \ldots, K$ and so by (3) above we have $$ \left\lvert f_n \left(x_i \right) - f \left(x_i \right) \right\rvert < \varepsilon \tag{6}$$ for each $i = 1, \ldots, K$.
Now if $x \in X$ is arbitrary, then $x \in U_{x_r}$ for some $r = 1, \ldots, K$ [Refer to (4) above.]. Then by (1) above we conclude that $$ f(x) \in \left( \ f \left( x_r \right) - \varepsilon, \, f \left( x_r \right) + \varepsilon \right), $$ that is, $$ f \left( x_r \right) - \varepsilon < f(x) < f \left( x_r \right) + \varepsilon. $$ which is equivalent to $$ \left\lvert f(x) - f \left( x_r \right) \right\rvert < \varepsilon. \tag{7} $$
So by (6) and (7) above, we conclude that, for all $x \in X$ and for every natural number $n > N$, we have $$ \left\lvert f_n(x) - f(x) \right\rvert \leq $$
Is what I've done so far correct? If so, is it going to lead us to our desired proof? If yes, then how to proceed from here? How to complete the proof?
Fix $\epsilon>0$. Let $U_n=\{x\in X: f(x)-f_n(x)<\epsilon\}$. Then $U_n$ is open because $f$ and $f_n$ are continuous. Furthermore, $\cup_{n=1}^\infty U_n=X$. This is because for each $x\in X$, there exists $n$ such that $f(x)-f_n(x)<\epsilon$. Note also that if $m<n$ and if $x\in U_m$, then since $f_m(x)\leqslant f_n(x)$, $$f(x)-f_n(x) \leqslant f(x)-f_m(x)<\epsilon.$$ Therefore $U_m\subset U_n$ when $m<n$.
Therefore $(U_n)_{n=1}^\infty$ is an open cover of $X$. By compactness, there is an open subcover, say $(U_{n_1}, \ldots, U_{n_t})$ with $n_1<\ldots <n_t$. But nowby the last sentence of the previous paragraph, $U_{n_1}\subset \ldots \subset U_{n_t}$. Therefore $$X\subset \cup_{i=1}^t U_{n_i}=U_{n_t}\subset \cap_{n\geqslant n_t}U_n.$$ Therefore for all $n\geqslant n_t$ and all $x\in X$, $f(x)-f_n(x)<\epsilon$. Since $f(x)-f_n(x)\geqslant 0$ for all $n$, we deduce that $\|f-f_n\|_\infty\leqslant \epsilon$ for all $n\geqslant n_t$.