Function defined as the supremum of a sequence of functions - epsilon-delta proofs I'm really confused with working with sequences of functions and their sup.

Let A be a non-empty subset of $\mathbb{R}$. For all $n \in \mathbb{N}$ and let $f_n : A \to\mathbb{R}$ be a continuous function. 

Suppose that for all $x \in A$ the sequence $f_1(x), f_2(x)...$ is bounded.

Define the function $g : A \to\mathbb{R}$ by: $g(x)=\sup\{f_1(x), f_2(x)...\}$

Then if we have some $a \in A$ and $\epsilon\gt0$, how would one go about proving that there is a $\delta\gt0$ such that, for all $x\in A$ with $|x-a|\lt\delta$ one has that $g(a)-g(x)\lt \epsilon$ ?

Additionally, how can we show whether the function $g$ is continuous?

I'm trying to do a standard epsilon-delta proof for the first part, but can't seem to introduce an $|x-a|$ term into the working.
Hint:
$$g(x) - g(a) \leqslant \sup g(x) - \inf g(a).$$
Use the fact:  $\sup A-\inf B=\sup(A-B).$
Hint:
$$g(x) - g(a) \leqslant \sup g(x) - \inf g(a).$$
Use the fact:  $\sup A-\inf B=\sup(A-B).$