Let $A$ be a metric space, and $U_{b}(A)$ the set of all bounded and uniformly continuous functions $g:A\rightarrow \mathbb{F}$.
Let $||\cdot||$ be the sup norm with $||g||=\sup_{x\in A}|g(x)|$.
How do I prove that $U_b(A)$ is a Banach space?
I know that being a Banach space means that $U_b(A)$ given the sup norm must be complet, i.e. each Cauchy sequence of functions $(g_n)_n$ must converge.
Further the definition of uniform continuity is: for every $\epsilon$ there exists a $\delta>0$ such that $|g(x)-g(y)|<\epsilon$ for all $x,y\in A$ s.t. $d(x,y)<\delta$.
I was thinking (assuming that $\mathbb{F}$ is complete), that we can fix a $a\in A$ and say that $(g_n(a))_n$ converges to a limit $g(a)$ in $\mathbb{F}$, if $(g_n)_n$ is Cauchy in $U_b(A)$. Then we can take $g$ to be the limit of $(g_n)_n$.
Is this a correct way of thinking? How do I finish my proof and where do I use the uniform continuity?
EDIT:
Uniform limit:
Let $\epsilon > 0$. Let $N$ s.t. for $n,m \geq N$ we have $\|g_n-g_m\| < \epsilon$. Then for all $n \geq N$
$$ |g_n(x) - g(x)| = \lim_{m \to \infty} |g_n(x) - g_m(x)| \leq \epsilon $$
for all $x$ and so $\|g_n- g\| \leq \epsilon$.
EDIT 2:
I am having difficulties showing the uniform continuity of $g$. Any and all help would be greatly appreciated!