Suppose we have functions $f_n$ with $f_n(x) \to g$ as $x \to 0$. I need to show that this convergence is uniformly in $n$ and I want to do it by proof by contradiction. But I have problems with the contradiction assumption.
As in, "Suppose the convergence is not uniform. Then ..." what to put here? There is something like: "Then for every $\epsilon > 0$ there exists a subsequence $n_j$ such that $|f_n(x)-g| \geq \epsilon$..." But I don't know.
Saying "$f_n(x)\to f_n(0)$ uniformly" doesn't makes many sense because the "limit" depends of $n$.
Even reinterpreting as "$f_n(x) - f_n(0)\to 0$ uniformly", is false: take $f_n(x) = n|x|$.
EDIT:
The uniform convergence means $$\forall\epsilon>0\ \exists\delta_\epsilon>0\ \forall x\in(-\delta_\epsilon,\delta_\epsilon)\ \forall n\in\Bbb N:\ |f_n(x) - g|<\epsilon.$$ The negation of this is: $$\exists\epsilon>0\ \forall \delta>0\ \exists x_\delta\in(-\delta,\delta)\ \exists n_\delta\in\Bbb N :\ |f_{n_\delta}(x_\delta) - g|\ge\epsilon.$$
And remember again: $g$ is a constant. The difference isn't $|\cdots - g(x)|$.