Proving Cauchy criterion for a sequence of functions. 
Proof.
(=>): First, we assume that $(f_n)$ converges uniformly on $A$, say to $f(x)$. Let $\epsilon > 0$ be given and choose and $N$ so that for all $n > N$ and all $x \in A$ one has $|f_n(x) - f(x)| < \epsilon $.
Now, Take $n_0 = N$.  Let $n \in \mathbb{N}$ be arbitrary. Clearly, $n+m > n_0$ as $m > m_0$ and so
$$ |f_{m+n}(x) - f_m(x) | \leq |f_{m+n} - f | + |f-f_m| < 2 \epsilon $$
Since this is true for any $x \in A$, then we are done.
(<=): Suppose now that the Cauchy Criterion is satisfied. Let $\epsilon > 0$. Let $x \in A$ and Pick $n_0$ so that if $m > n_0$ then (with $n=1$)
$$ |f_m - f | = | f_m - f_{m+1} | + |f_{m+1} - f | < 2 \epsilon  $$
and thus $f_m $ converges uniformly to $f$ and we done.
IS this enough for a solution? Any constructive criticism is extremely welcome!
 A: As was noted in the comments, the forward direction is fine apart from a the typo of $m_0$ for $n_0$. For the other direction, however, you’re going to need to use that Cauchy criterion to construct a function $f$ and then show that the functions converge uniformly on $\mathbf{A}$ to that function $f$.
We’re assuming that for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $|f_{n+m}(x)-f_m(x)|<\epsilon$ whenever $m>n_\epsilon$, $n\in\Bbb N$, and $x\in\mathbf{A}$. This is clearly equivalent to saying that for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $|f_n(x)-f_m(x)|<\epsilon$ whenever $m,n>n_\epsilon$ and $x\in\mathbf{A}$, which means that $\langle f_n(x):n\in\Bbb N\rangle$ is a Cauchy sequence in $\Bbb R$ (or possibly $\Bbb C$) for each $x\in\mathbf{A}$. Every Cauchy sequence in $\Bbb R$ (or $\Bbb C$) converges, so we can define a function $f$ on $\mathbf{A}$ by setting
$$f(x)=\lim_{n\to\infty}f_n(x)$$
for each $x\in\mathbf{A}$.
Now that we have $f$, we can check that $\langle f_n:n\in\Bbb N\rangle$ converges uniformly on $\mathbf{A}$ to $f$. We’ll use the following lemma.

Lemma. Let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb R$ (or $\Bbb C$) converging to $x$, and suppose that $\epsilon>0$ and $n_0\in\Bbb N$ are such that $|x_n-x_m|<\epsilon$ whenever $n,m>n_0$; then $|x_n-x|\le\epsilon$ for each $n>n_0$.
Proof. Suppose that $|x_n-x|>\epsilon$ for some $n>n_0$, and let $\delta=|x_n-x|-\epsilon>0$. The sequence converges to $x$, so there is an $m>n_0$ such that $|x_m-x|<\delta$. But then $$|x-x_n|\le|x-x_m|+|x_m-x_n|<\delta+\epsilon=|x-x_n|\;,$$ which is absurd. Thus, $|x_n-x|\le\epsilon$ for all $n>n_0$. $\dashv$

Let $\epsilon>0$, and let $n_0\in\Bbb N$ be such that $|f_n(x)-f_m(x)|<\frac\epsilon2$ whenever $m,n>n_0$ and $x\in\mathbf{A}$. It follows immediately from the lemma that $|f_n(x)-f(x)|\le\frac\epsilon2<\epsilon$ for all $n>n_\epsilon$ and $x\in\mathbf{A}$, i.e., that $\langle f_n:n\in\Bbb N\rangle$ converges uniformly to $f$.
