Proving that $(f_n)$ converges uniformly on $A$. 
try:
Let $\epsilon > 0$. $(a_n)$ convergent means that $(a_n)$ is Cauchy sequence so that we can find $N$ with $n,m > N$ and satisfying $|a_n - a_m| < \epsilon$. By definition of sup, we have that
$$ |f_n(x) - f_m(x)| \leq \sup_{x \in A} |f_n(x) - f_m(x)| \leq |a_n-a_m| < \epsilon $$
so if we put $m=N+1$ and call $g(x) = f_{N+1}(x)$ then we observe that for all $n>N$ we obtain
$$ |f_n(x) - g(x) | < \epsilon $$
for all $x \in A$ meaning that $f_n $ converges uniformly to g(x) in A as was to be shown.
IS this correct?
 A: I assume in your question the $f_n$'s are functions $A \to \Bbb{R}$. But all that is needed is for the target space to be any complete metric space. Notice that part of your proof shows in particular shows that for every $x \in A$ the sequence $\{f_n(x)\}_{n \in \Bbb{N}}$ is a Cauchy sequence in $\Bbb{R}$. Since $\Bbb{R}$ is complete, this sequence of numbers has a limit. Call it $f(x)$. 
Thus, we have a pointwise limit $f: A \to \Bbb{R}$. I claim now that the convergence $f_n \to f$ is also uniform. For this, let $\epsilon > 0$ be given. Then, there is an $N$ such that for all $m,n > N$, and ALL $x \in A$,
\begin{align}
|f_n(x) - f_m(x)| < \epsilon.
\end{align}
Let $n \geq N$, and let $x \in A$; we shall now take the limit $m \to \infty$ in the above inequality: 
\begin{align}
\lim_{m \to \infty}|f_n(x) - f_m(x)| &\leq \lim_{m \to \infty} \epsilon\\
\left| \lim_{m \to \infty}(f_n(x) - f_m(x))\right| & \leq \epsilon \\
\left| f_n(x) - f(x)\right| & \leq \epsilon,
\end{align}
where in the second step, I moved the limit inside the absolute value because $|\cdot|$ is a continuous function, so we can pull limits in and out. 
