Generally when i want to show that $f_n$ convergence uniformly I use the Dini's Theorem or others to show that it doesn't but sometimes I have to go with the definition and show that $\sup\{| f_{n}(x)-f(x))|\}\to0$ as $n\to \infty$.
How exactly I work to show that, using $\sup\{|f_{n}(x)-f(x))|\}\geq| f_{n}(x)-f(x)|$ ?
I leave an example that I think I have to use the definition where I try to prove it.
Let $f_{n}(x)=(-1)^{n}\dfrac{x^{2}+n}{n^{2}},\:x\in[ 0,1 ],\: n\in \mathbb{N}$ then $f_{n}(x)\leq\dfrac{2n}{n^{2}}$ since $x\in[0,1]$ and $n\geq 1\Rightarrow|\sup\{ f_{n}(x)\}|\leq\dfrac{2n}{n^{2}}$ and $\dfrac{2n}{n^{2}}\to0\Rightarrow|\sup\{f_{n}(x)\}|\to0$.
If I understand what you are asking:
Let $f_n:E\to\mathbb{R}$,$\quad f:E\to\mathbb{R}$.
Uniform convergence: For any $\epsilon>0$, there exists $N$ such that $|f_n(x)-f(x)|<\epsilon$ for all $x\in E$ and $n\geq N$. Since this holds for all $x$ and $n$ does not depend on $x$, it follows that $\sup_{x\in E}|f_n(x)-f(x)|\leq \epsilon$.
Conversely, if there exists $N$ such that for all $n\geq N$, we have $\sup_{x\in E}|f_n(x)-f(x)|\leq \epsilon$, then for all $x\in E$, we have $$ |f_n(x)-f(x)|\leq \sup_{x\in E}|f_n(x)-f(x)|\leq \epsilon $$ so $f_n\to f$ uniformly.
As far as the example you provided, $f_n(x)=(-1)^n \frac{x^2+n}{n^2}$ for $x\in[0,1]$, in order to show $f_n \to 0$ uniformly, we observe that $|f_n(x)|=\frac{x^2+n}{n^2}\leq \frac{1+n}{n^2} \to 0$ as $n\to\infty$. The key is we obtained a bound that does not depend on $x$. Hence, when you choose $N$ such that $\frac{1+n}{n^2}<\epsilon$ for all $n\geq N$, it also follows that $|f_n(x)|<\epsilon$.
It's pretty easy to show that $\sup |f_n(x)|=\frac{1+n}{n^2}$, but in a harder problem, you really just need an upper bound that is independent of $x$ and converges to $0$.