Assume that we can find $\delta>0$, a subsequence $\{f_{n_k}\}$ and a sequence $\{x_{n_k}\}$ such that $f_{n_k}(x_{n_k})\geqslant \delta$ for all $k$. Bolzano-Weierstrass theorem ($[a,b]$ is compact) allows us to extract of $\{x_{n_k}\}$ a subsequence, denoted $\{t_j\}$, converging to some $t$. We have for all integers $n$ and $m$, denoting $g_k$ the sequence indexed by the integers appearing in $t_k$,
$$\delta\leqslant g_{m+n}(t_{m+n})\leqslant g_n(t_{m+n}).$$
Now we fix $n$, and take the $\limsup_{m\to +\infty}$. This gives for all integer $n$,
$$\delta\leqslant\limsup_{k\to +\infty}g_k(t_k)\leqslant g_n(t),$$
a contradiction.
In the second theorem, take $a=0, b=1$.
- If we don't assume $f$ continuous, consider $f_n(x):=x^n$.
- The same counter-example considering $(0,1)$.
Here's one that I rather like that uses equicontinuity and compactness as the engine for moving from pointwise to uniform convergence. For the sake of clarity, let's recall what equicontinuity means.
Definition: Let $X$ and $Y$ be metric spaces. A set $E$ of continuous functions from $X$ to $Y$ (i.e. $E \subseteq C(X,Y)$ ) is equicontinuous if for every $x \in X$ and every $\varepsilon>0$ there exists $\delta >0$ such that
$$
y \in X \text{ and } d_X(x,y) < \delta \Rightarrow d_Y(f(x),f(y)) < \varepsilon \text{ for all } f \in E.
$$
With this definition in hand, we can state the result.
Theorem: Let $X$ and $Y$ be metric spaces with $X$ compact. Suppose that for $n \in \mathbb{N}$ the function $f_n: X \to Y$ is continuous and that $\{f_n | n \in \mathbb{N}\}$ is equicontinuous. Let $f : X \to Y$ and suppose that $f_n \to f$ pointwise as $n \to \infty$. Then $f_n \to f$ uniformly as $n \to \infty$. In particular, this means that $f$ is continuous.
Proof:
Let $\varepsilon >0$. By the equicontinuity property, for each $x \in X$ we can find $\delta_x >0$ such that
$$
y \in B_X(x,\delta_x) \Rightarrow d_Y(f_n(x),f_n(y)) < \frac{\varepsilon}{4} \text{ for all } n \in \mathbb{N}.
$$
The collection of open balls $\{B_X(x,\delta_x)\}_{x \in X}$ is an open cover of $X$. The compactness of $X$ allows us to find a finite subcover, namely $x_1,\dotsc,x_k \in X$ such that $X = \bigcup_{i=1}^k B_X(x_i,\delta_{x_i})$. Due to the pointwise convergence $f_n \to f$ (or really the pointwise Cauchy property), for each $i=1,\dotsc,k$ we can choose $N_i \in \mathbb{N}$ such that
$$
m,n \ge N_i \Rightarrow d_Y(f_n(x_i),f_m(x_i)) < \frac{\varepsilon}{4}.
$$
Set $N = \max \{N_1,\dotsc,N_m\} \in \mathbb{N}$. For each $x \in X$ there exists $i\in \{1,\dotsc,k\}$ such that $x \in B(x_i,\delta_{x_i})$. Then for $m \ge n \ge N$ we have the estimate
$$
d_Y(f_m(x),f_n(x)) \le d_Y(f_m(x),f_m(x_i)) + d_Y(f_m(x_i),f_n(x_i)) + d_Y(f_n(x_i),f_n(x)) \\
< \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} = \frac{3\varepsilon}{4}.
$$
Hence for $n \ge N$,
$$
d_Y(f(x),f_n(x)) = \lim_{m \to \infty} d_Y(f_m(x),f_n(x)) \le \frac{3\varepsilon}{4}.
$$
This holds for all $x \in X$, so
$$
n \ge N \Rightarrow \sup_{x \in X} d_Y(f(x),f_n(x)) \le \frac{3\varepsilon}{4} < \varepsilon,
$$
and we deduce that $f_n \to f$ uniformly as $n \to \infty$.
Best Answer
Let $A=\{x: f_n(x) \leq f_{n+1}(x) \forall n\}$ and $B=\{x: f_n(x) \geq f_{n+1}(x) \forall n\}$. Note that $A$ and $B$ are closed sets and hence they are also compact. Also $A \cup B=X$. $f_n \to f$ uniformly on each of these sets. Given $\epsilon >0$ there exist $n_1, n_2$ such that $|f_n(x)-f(x)| <\epsilon$ for all $x \in A$ for all $n >n_1$ and $|f_n(x)-f(x)| <\epsilon$ for all $x \in B$ for all $n >n_1$. Let $n_0=\max \{n_1,n_2\}$. Then $|f_n(x)-f(x)| <\epsilon$ for all $x \in X$ for all $n >n_0$.