Is this enough for uniform convergence?

Question

Let $f_n$ be a sequence of continues functions from $(-a,a)$ to $\mathbb{C}$ such that $\lim_{n \to \infty} f_n(x)=0$ does this mean $f_n(x)$ is uniformly convergent on $[-\epsilon,\epsilon]$ whenever $\epsilon\in(0,a)$? Here's my attempt to prove this:
$|f_n(x)|$ is continues, so it has maximum in $[-\epsilon,\epsilon]$. Let's denote it by $M_n$. $\lim_{n \to \infty} f_n(x)=0$ thus $\lim_{n \to \infty}M_n=0$. That finishes the proof since $M_n$ can be arbitrarily small and is the upper bound for $|f_n(x)-0|$.

Answer 1

"$\lim_{n \to \infty} f_n(x)=0$ thus $\lim_{n \to \infty}M_n=0$." This part is not correct, because the minima are not reached at the same point. Pointwise convergence of $f$ to $0$ does not imply that for any sequence $x_n\to 0$, we have $f_n(x_n) \to 0$.

In fact, the result is false. Let $f_n$ be a continuous piecewise affine function defined by : $$\forall n\in \mathbb N,\forall x\in\mathbb R, f_n(x) = \left\{\begin{array}{cl} nx & \text{if } x\in [0,1/n] \\ 2 - nx &\text{if } x\in [1/n,2/n]\\ 0& \text{else} \end{array}\right.$$

Then, it is easy to see that for any $x\in \mathbb R$, $\lim_{n\to +\infty} f_n(x) = 0$. However, $(f_n)$ does not converge uniformly on any neighborhood of $0$. Indeed, for any $\epsilon>0$, we have: $$\lim_{n\to +\infty} \max_{[-\epsilon ,\epsilon]}|f_n| = 1$$