Let us suppose to have a series ${f_N} ({N\in\mathbb{N}})$ such that, for each $N$, $f_N: \mathbb{R}\mapsto\mathbb{R}$.
Besides, I know that $f_N$ is continuous for all $N\in\mathbb{N}$ and $f=L^2\text{-}\lim_\limits{N\to\infty}f_N$.
I would like to show that $f_N$ uniformly converges to $f$. First, consider the sequence $\left(f_N\right)_{N\in\mathbb{N}}$.
I am told that in order to show that $f_N$ uniformly converges to $f$, it suffices to prove that:
$$\text{"a subsequence of }(f_N)_{N\in\mathbb{N}} \text{ converges uniformly to } f"$$
Why is that true? Could you please explain such a statement/give a reference about that?
It is not true. Short version: you can join a uniformly convergent and a pointwise but not uniformly convergent sequence with the same limit in one sequence and generate a counterexample.
Now let us explicitly write down a counterexample. We choose $$ f_n(x)= \sqrt{x^2+\frac 1n} \quad\mbox{for even }n\qquad\mbox{and}\quad f_n(x)=|x|+\frac xn \quad\mbox{for uneven }n$$ Then we have $f_n(x)\to|x|=:f(x)$ for all $x\in\mathbb R$. The convergence $f_{2n}\to f$ is also uniform. However, $$\sup\{f_n(x)-f(x):x\in\mathbb R\}=\infty,$$ So the whole sequence doesn‘t converge uniformly.