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I have proved the following and I would like to know if I have made any mistakes:

A sequence of functions $(f_n)$ defined on a set $A\subseteq\mathbb{R}$ converges uniformly on $A$ if and only if for every $\varepsilon>0$ there exists an $N\in\mathbb{N}$ such that $|f_n(x)-f_m(x)|<\varepsilon$ whenever $m, n\geq N$ and $x\in A$.

My proof (NOTE: edited leftward implication according to answer below):

$\Rightarrow$ Let $\varepsilon>0$: then by hypothesis there exists a limit function $f$ and $N\in\mathbb{N}$ such that $|f_n(x)-f(x)|<\frac{\varepsilon}{2}$ for all $n\geq N$ and $x\in A$ so if we pick $m\geq N$ we have also that $|f_m(x)-f(x)|<\frac{\varepsilon}{2}$ for all $m\geq N$ and $x\in A$ so $|f_n(x)-f_m(x)|=|f_n(x)-f(x)+f(x)-f_m(x)|\leq |f_n(x)-f(x)|+|f(x)-f_m(x)|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$ for all $m,n\geq N$ and $x\in A$, as desired.

$\Leftarrow$ Fix $x\in A$: then $(f_n(x))_{n\in\mathbb{N}}$ is a Cauchy sequence of real numbers which is convergent by the Cauchy Criterion for sequences so the function $f:A\to\mathbb{R}, f(x):= \lim_{n\to\infty} f_n(x)$ is well defined for every $x\in A$. Now, let $\varepsilon>0$: by hypothesis we have that there is $N\in\mathbb{N}$ such that $|f_n(x)-f_m(x)|<\frac{\varepsilon}{2}$ for all $m,n\geq N$ and $x\in A$ so $|f(x)-f_n(x)|=\lim_{m\to\infty}|f_m(x)-f_n(x)|\leq\frac{\varepsilon}{2}<\varepsilon$ for $n\geq N$ and $x\in A$, as desired.

Thank you.

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In the $\Leftarrow$ direction, note that your $N$ may depend on $x$. You write (red additions by me):

So, if $x\in A$ and $\varepsilon>0$, there exists $N\color{red}{(x,\varepsilon)}\in\mathbb{N}$ such that $|f_n(x)-f(x)|<\frac{\varepsilon}{2}$ for all $n\geq N$ (since $\lim_{n\to\infty} f_n(x)=f(x)$ for $x\in A$).

You have to show that there is an $N \in \mathbb N$ that does the trick for any $x \in A$. Let $\varepsilon > 0$, by the uniform Cauchy condition there is $N \in \mathbb N$ such that $$ \left|f_n(x)- f_m(x) \right| < \varepsilon, \qquad n, m \ge N, x \in A. $$ (Note that $N$ is independent of $x$). Now let $x \in A$. As $f_m(x) \to f(x)$ for $m \to \infty$, we have $$ \left|f_n(x)- f(x) \right| = \lim_{m \to \infty} \left|f_n(x)- f_m(x) \right|\le \varepsilon $$ That is $$ \left|f_n(x)- f(x) \right| \le \varepsilon, \qquad n \ge N, x \in A.$$

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