I came across this problem when I reading this undergrad analysis book, and I think this problem is a little bit challenging unless I was overdoing it. Here is my proof:
First we have $A=\bigcup_{\delta>0} A(\delta)$ is bounded above by $b$ by the construction of $A(\delta)$, and it's nonempty because it contains $a$. So there exists $c=l.u.b.(A)$, of course $c\leq b$ since $b$ is an upper bound.
Some fact:
(1) if $u \in A(\delta)\subset A$, then for any $v\in[a,u)$, $v$ is also in the same $A(\delta)$, therefore in $A$. This is just saying that if there is any $u\in A$, then $[a,u]\subset A$.
Proof: play with the definition.(omitted)
(2)For fixed $\epsilon$, if $s\in A$ where A is induced by $\epsilon$, then $s\in A$ where A is induced by $\frac{\epsilon}{2}$. Proof: rescale $\epsilon$.
Since function $f$ is continuous at $c$, for any $\epsilon >0$,there exists $\delta_0>0$ such that for any $y$ that satisfies $|y-c|\leq\delta_0$ we have $|f(c)-f(y)|\leq \epsilon$. Now we pick this very $\delta_0$, since $c$ is the l.u.b.(A) so there exists $s\in A$ such that $c-\delta_0\leq s\leq c$, since $s\in A=\bigcup_{\delta>0}A(\delta)$, so there exists $\delta_1$ such that $s\in A(\delta_1)$.
Now we want to show that $c$ is also in $A$:
Fact (1) tells us $[a,s]\subset A(\delta_1)$, let $\delta=min\{\delta_0, \delta_1\}$. Then for any $x,t \in [a, c]$ with $|x-t|<\delta$,
case1: If $x,t$ are both in $[a,s]$, then $|f(x)-f(t)|<\epsilon$ because $f$ is uniformly continuous on $[a,s]$.
case2: If $x,t$ are both in $[s, c]$, then $|f(x)-f(t)|\leq |f(x)-f(c)|+|f(c)-f(t)|<\epsilon + \epsilon=2\epsilon$.
case3: If one is in $[a,s]$ and the other is in $[s,c]$. WLOG, let $x\in [a,s]$ and $t\in [s,c]$, then $|f(x)-f(t)|\leq |f(x)-f(s)|+|f(s)-f(c)|+|f(c)-f(t)|\leq \epsilon + \epsilon +\epsilon =3\epsilon$.
Anyway, we can use fact (2) to rescale and get a uniform $\epsilon$. So we've shown $c$ is also in $A$. Now if we assume $c<b$, then we can repeat the above argument to show $c+\delta_0$ is also is also in $A$ (use $c$ as the intermediate point), contradicting the fact the $c$ is the l.u.b.(A). So the only possible way is that $c=b$ and $c+\delta_0$ is out of the bound of domain $[a,b]$. Therefore $c=b\in A$, done.
P.S.: I've carefully checked my reasoning and it should be right up to some detail work, e.g. rescaling issue. And I'm pretty sure there is a better way to put all these together nicely, but I'm out of time so...
And now for something slightly different:
Since $f,g$ are bounded on $A$, their ranges lie in some compact set
$[-B,B]$. Since $[-B,B]^2$ is compact, we see that multiplication $\cdot : [-B,B]^2 \to \mathbb{R}$ is uniformly continuous.
Since $f,g$ are uniformly continuous separately, we see that the map
$p(x) = (f(x),g(x))$ is also uniformly continuous.
Since the composition of uniformly continuous maps is again uniformly continuous (this follows almost immediately from the definition), we see that
$\cdot \circ p$ is uniformly continuous.
Best Answer
I'm not sure if your post is indicating that you know this, but there is a quick proof using a relatively well known fact: the sum of two continuous functions is continuous. Assume $f+g$ is continuous. Then since $f$ is continuous, $-f$ is continuous. Therefore $(f+g)+(-f)=g$ is continuous, contradiction.
If you don't know this fact already, I would recommend trying to prove it for yourself. The crucial step is to consider $\delta=\max{(\delta_f,\delta_g)}$. You can actually use the proof of the above statement to prove your question by contradiction. As you've noted, the fact that $\delta$ can be any real number can be problematic. However, you do have a statement that is true for all $\delta$, namely
$$\exists\epsilon_g\forall\delta_g\exists x_g\text{ such that }|x_g-a|<\delta_g\wedge |g(x_g)-g(a)|>\epsilon_g$$
This fact is the one that you want to leverage to get your result. I think by contradiction would be the best way to proceed. So, you assume $$(\forall\epsilon)(\exists\delta)(\forall x)(|x-a|<\delta\Rightarrow|f(x)+g(x)-f(a)+g(a)|<\epsilon)$$ and you have that $$(\forall\epsilon')(\exists\delta')(\forall x)(|x-a|<\delta'\Rightarrow|f(x)-f(a)|<\epsilon')$$ so therefore $$|g(x)-g(a)|=|f(x)+g(x)-(f(a)+g(a))-(f(x)-f(a))|\leq \epsilon+\epsilon'$$ by the triangle inequality, which contradicts that $g$ is discontinuous once you do a bit of tweaking of $\delta$'s and $\epsilon$'s.