Let $(X,d)$ be a compact metric space, $f_n : X \to \mathbb{R}$ a sequence of continuous functions, $f_{\infty} : X \to \mathbb{R}$ a continuous function. Now suppose that $f_n \to f$ pointwise, and that for each $x \in X$ the sequence (of real numbers) $\{ f_n (x) \}$ is monotonic (not necessarily all monotonic in the same way, there could exist $x$ such that $f_n (x)$ is monotonically increasing and $y$ such that $f_n (y)$ is monotonically decreasing). Then $f_n \to f$ uniformly. I think i found a simple proof of this theorem but i'm not sure it works.
Let's assume by way of contradiction that the convergence is not uniform. Therefore there exists a positive real number $\rho > 0$ such that $\sup_{x \in X} |f_n(x)-f_{\infty}(x)| \geq 2 \rho$ for infinitely many $n \in \mathbb{N}$. For each one of these $n$'s, choose $x_n \in X$ such that $|f_n (x_n) - f_{\infty} (x_n)| \geq \rho$. Because we have infinitely many of these $n$'s and each $x_n$ can be either such that $f_N (x_n)$ is monotonically increasing or monotonically decreasing (wrt to $N$), there are either infinitely many of these $n$'s such that $f_N (x_n)$ is monotonically increasing or infinitely many of these $n$'s such that $f_N (x_n)$ is monotonically decreasing. Suppose the first one is true (the other case is identical). Then there exists a strictly increasing sequence $\{n_k\}_{k=1}^\infty \subseteq \mathbb{N}$ such that for each $k \geq 1$ $|f_{\infty} (x_{n_k}) - f_{n_k} (x_{n_{k}})| \geq \rho$ and $f_N(x_{n_k})$ is monotonically increasing. Now, because $f_N (x_{n_k}) \to f_{\infty} (x_{n_k})$ as $N \to \infty$ and $f_N(x_{n_k})$ is monotonically increasing, we get $f_{n_k}(x_{n_k}) \leq f_{\infty} (x_{n_k})$, and so $f_{\infty} (x_{n_k}) - f_{n_k} (x_{n_{k}}) \geq \rho \iff f_{\infty} (x_{n_k}) \geq f_{n_k} (x_{n_k}) + \rho$. Now, this, together with the monotonicity, implies that for each fixed positive integer $N$ we have $f_{\infty} (x_{n_k}) \geq f_{N} (x_{n_k}) + \rho$ for big enough $k$. Now, $X$ is compact, therefore ${x_{n_k}}$ has a convergent subsequence $\{ x_{n_{k_t}}\}$ converging to a point $x_0 \in X$. So we get, for $t$ big enough, $f_{\infty} (x_{n_{k_t}}) \geq f_{N} (x_{n_{k_t}}) + \rho$. Taking the limit as $t\to \infty$, because of the continuity we get $f_{\infty} (x_0) \geq f_N (x_0) + \rho$ for all $N \in \mathbb{N}$. Taking the limit as $N \to \infty$, we get the contradiction $f_{\infty} (x_0) \geq f_{\infty}(x_0) + \rho$
The proof is correct even when $X$ is any sequentially compact space. Good job!