I hope the following proof is correct.
First, let us show that given $\varepsilon >0$, one can find $N^+\in\mathbb N$ such that $$\forall n\geq N^+\,\forall t\in [a,b]\;:\; f_n(t)\leq f(t)+\varepsilon\, . $$
Since $f$ is uniformly continuous on $[a,b]$, one can find $a=a_0<\dots <a_K=b$ such that for $i=0,\dots ,K-1$, it holds that $$\forall t\in [a_i,a_{i+1}]\;:\;f(t)\geq \max(f(a_i), f(a_{i+1}))-\frac\varepsilon2\cdot$$
Choose $N^+$ such that if $n\geq N^+$, then $\vert f_n(a_i)-f(a_i)\vert\leq\frac\varepsilon2$ for $i=0,\dots ,K$. Since $f_n$ is convex, you then have for $i=0,\dots ,K-1$ and $t\in [a_i,a_{i+1}]$:
$$f_n(t)\leq \max(f_n(a_i), f_n(a_{i+1}))\leq \max(f(a_i), f(a_{i+1}))+\frac\varepsilon2\cdot$$
If you compare with the previous inequality, this gives that if $n\geq N^+$, then $f_n(t)\leq f(t)+\varepsilon$ for all $t\in[a,b]$, as required.
Now, let us show that given $\varepsilon >0$, one can find $N^-$ such that
$$\forall n\geq N^+\,\forall t\in [a,b]\;:\; f_n(t)\geq f(t)-\varepsilon\, . $$
Towards a contradiction, assume that this is not the case. Then one can find a sequence $(t_k)\subset [a,b]$ and a subsequence $f_{n_k}$ of $(f_n)$ such that $$\forall k\in\mathbb N\; :\; f_{n_k}(t_k)< f(t_k)-\varepsilon\, . $$
By compactness, we may assume that the sequence $(t_k)$ is convergent; and there is no loss of generality in assuming that $t_k\to a$ or $t_k\to b$. Let's assume for example that
$$ t_k\to a\, .$$
For each $k$, let $\psi_k$ be the unique affine function such that $f_{n_k}(a)-\psi_k(a)=0=f_{n_k}(b)-\psi_k(b)$. Since $f_{n_k}(a)\to f(a)$ and $f_{n_k}(b)\to f(b)$, one can check easily that $(\psi_k)$ converges uniformly on $[a,b]$ to the unique affine function $\psi$ such that $f(a)-\psi(a)=0=f(b)-\psi(b)$. Replacing $f_{n_k}$ by $f_{n_k}-\psi_k$ and $f$ by $f-\psi$, it follows that without loss of generality, we may assume that
$$\forall k\in\mathbb N\; :\; f_{n_k}(a)=0=f_{n_k}(b)\, .$$
Then $f(a)=0=f(b)$. In particular, since $f$ is continuous, we may choose $\delta>0$ such that $f(t)\geq-\frac\varepsilon3$ on $[0,\delta]$. We may of course assume that $\delta\leq \frac{a+b}2$, and since $t_k\to a$ we may also assume that $t_k\in[0,\delta]$ for all $k$.
Now, consider for each $k$ the function $\varphi_k$ defined as follows : $\varphi_k(a)=0=\varphi_k(b)$, $\varphi_k(t_k)=f(t_k)-\varepsilon$, and $\varphi_k$ is affine on $[a,t_k]$ and on $[t_k,b]$. Since $f_{n_k}$ is convex and not greater than $\varphi_k$ at the "node" points, we have $f_{n_k}\leq\varphi_k$ on $[a,b]$. In particular,
$$f_{n_k}(\delta)\leq \varphi_k(\delta)=\varphi_k(t_k)+\frac{\varphi_k(b)-\varphi(t_k)}{b-t_k}\, (\delta-t_k)\, ;$$
that is,
$$f_{n_k}(\delta)\leq f(t_k)-\varepsilon+\frac{\varepsilon-f(t_k) }{b-t_k }\, (\delta- t_k)\cdot $$
Since $f_{n_k}(\delta)\to f(\delta)$, $t_k\to a
$ and $f(t_k)\to f(a)=0$, it follows that
$$f(\delta)\leq -\varepsilon +\frac{\delta-a}{b-a}\,\varepsilon\leq -\frac\varepsilon2\, , $$
where we have used the fact that $\delta\leq\frac{a+b}2\cdot$ Since $f(\delta)\geq -\frac\varepsilon3$, this is a contradiction.
Hence, the integer $N^-$ does exist, and the proof is complete.
related problem: (I), (II), (III). Here is a systematic technique. In order to find $\sup_{0\leq x\leq 1} |f_n(x)-f(x)| $, you need to maximize the function $\Big|\frac{x}{2n^2x^2+8}\Big|$ over the interval $[0,1]$. Now, let
$$ g(x)=\frac{x}{2n^2x^2+8} \implies g'(x) = \frac{4-n^2 x^2}{(2n^2x^2+8)^2}=0 \implies x=\frac{2}{n}$$
gives the max of the function $g(x)$ which is $g(2/n)=1/8n$. You can check this by checking the sign of $g''(x)$ which should be $< 0$. Hence we have
$$ \sup_{0\leq x\leq 1} |f_n(x)-f(x)|= \sup_{0\leq x\leq 1} \Big|\frac{x}{2n^2x^2+8}\Big|=\frac{1}{8n} < \epsilon. $$
Best Answer
Let $\varepsilon>0$ be given, and set $\delta=\min\left[\frac{\varepsilon}{3}, \frac{\varepsilon}{3L}\right]$. Since the collection of open balls $\mathcal{B}: = \{B(\, x, \delta) : x \in [a,b] \}$ is a cover for $[a,b]$ we may find a finite subcover, say $\{B(\,x_1, \delta), \, \ldots, \, B(\,x_M, \delta)\}$ (Heine-Borel Theorem). Since $f_n$ converges pointwise on $[a,b]$, for each point $x_j \: \left(\,j=1,\ldots, M \right)$ we may find a positive integer $N_j$ so that \begin{equation} \left|\, f_n(x_j) - f_m(x_j) \right| < \frac{\varepsilon}{3} \text{ whenever } n, m \geq N_j \,. \end{equation} Setting $N = \max [N_1, \ldots, N_M]$ shows that
\begin{aligned} \left|\,f_n(x)- f_m(x) \right| & \leq \left| \,f_n (x)- f_n(x_j) \right| + \left|\, f_n (x_j)- f_m(x_j) \right| + \left|\, f_m(x_j)- f_m(x) \right| \\ & < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon \; \: \text{ whenever } \, n,m \geq N \text{ and } x \in [a,b] . \end{aligned}
Since $\mathbb{R}$ is complete, it follows that the sequence of functions $\{\,f_n\}_{n=1}^\infty$ converges uniformly on $[a,b]$ (Cauchy Criterion).
I prefer uniform convergence first, ask $\,f$ questions later -_-.