Let $\varepsilon>0$ be given, and define the set $[m]:=\{\,j \in \mathbb{N} : 1 \leq j \leq m\}$.
For each $j \in [m]$, there is a positive number $\delta_j$ so that $|\,f_j(\mathbf{x})-f_j(\mathbf{a})|<\frac{1}{2}\varepsilon$ whenever $d(\mathbf{x,a})<\delta_j$ (Principle of Finite Choice).
We set $\delta=\min \{\delta_1, \delta_2, \ldots, \delta_m\}$.
Now we have that $f_j(\mathbf{a})-\frac{\varepsilon}{2}<f_j(\mathbf{x})<f_j(\mathbf{a})+\frac{\varepsilon}{2}$ whenever $d(\mathbf{x,a})<\delta$, and $j \in [m] .$
Since the closed ball $\bar{B}(\mathbf{a},\delta)$ is a compact subset of $\mathbb{R}^n$, we know that the set
\begin{equation}
\displaystyle \overset{m}{\underset{j=1}{\bigcup}} \, f_j(\bar{B}(\mathbf{a},\delta))
\end{equation}
is a compact subset of $\mathbb{R}$. Thus $f(\mathbf{x}):= \max \{\, f_1(\mathbf{x}), f_2(\mathbf{x}), \ldots, f_m(\mathbf{x})\}$ exists for all $\mathbf{x} \in \bar{B}(\mathbf{a},\delta)$.
Furthermore, for each $\mathbf{x} \in B(\mathbf{a},\delta)$ there is $j(\mathbf{x}) \in [m]$ such that
\begin{equation}
f(\mathbf{x})-\frac{\varepsilon}{2} < f_{j(\mathbf{x})}(\mathbf{x}) \leq f(\mathbf{x}), \text{ and so we have }
\\
f(\mathbf{x}) < f_{j(\mathbf{x})}(\mathbf{x}) + \frac{\varepsilon}{2} < f_{j(\mathbf{x})}(\mathbf{a}) + \varepsilon \leq f(\mathbf{a})+\varepsilon.
\end{equation}
Moreover, there is $j \in [m]$ so that $f(\mathbf{a})-\varepsilon < f_j(\mathbf{a})-\frac{\varepsilon}{2} < f_j(\mathbf{x}) \leq f(\mathbf{x})$.
We combine to conclude that $f(\mathbf{a})-\varepsilon < f(\mathbf{x}) < f(\mathbf{a}) + \varepsilon$ whenever $d(\mathbf{x,a})<\delta$. Therefore $f(\mathbf{x})=\max\{\, f_1(\mathbf{x}), f_2(\mathbf{x}), \ldots, f_m(\mathbf{x})\}$ is continuous at $\mathbf{a} \in S$.
Notice that while the particular family of functions is finite, a similar argument shows that $f(\mathbf{x}):=\sup_{\alpha \in A}\{\,f_{\alpha}(\mathbf{x})\}$ is continuous, for any equicontinuous family (countable or uncountable) of uniformly bounded functions. To be clear, in the case of an infinite family of functions the two assumptions, "uniformly bounded and equicontinuous" are sufficient. It should be clear from the proof above how these two assumptions are essentially given in the case of a finite family of continuous functions on a compact subset of $\mathbb{R}^n$.
Note: $j: B(\mathbf{a},\delta) \longrightarrow [m]$ $(*)$ is a choice function that assigns to each $\mathbf{x} \in B(\mathbf{a},\delta)$ an integer $j(\mathbf{x}) \in [m]$ in accordance with this characterization of supremum of subset of $\mathbb{R}$ which I rather mention than $\max$ since $\sup$ is relevant in the previously alluded to case of an infinite family (ie, $\alpha: B(\mathbf{a},\delta) \longrightarrow A$) and it is also what I had in mind when I originally wrote the preceding argument.
$(*)$ We could also consider the set $\left\{\{f_1(\mathbf x),f_2(\mathbf x),\ldots, f_m(\mathbf x)\}:\mathbf x \in B(\mathbf a, \delta) \right\}$ to be the domain of $j$ and the set $\bigcup_{\mathbf x \in B(\mathbf a, \delta)}\{f_1(\mathbf x),f_2(\mathbf x),\ldots, f_m(\mathbf x)\}$ to be the codomain of $j$.
D. S. Bridges and L. S. Vita; "Techniques of Constructive Analysis"
So given $\varepsilon>0$, we consider the set $S_{\varepsilon}:=\left\{(\mathbf{x},y) \in B(\mathbf{a},\delta) \times [m]: f(\mathbf{x})-\frac{1}{2}\varepsilon<f_y(\mathbf{x})\right\}$ and we know that for each $\mathbf{x} \in B(\mathbf{a},\delta)$ there exists $y \in [m]$ such that $(\mathbf{x},y) \in S_{\varepsilon}$ since $f(\mathbf{x})=\sup\{\, f_1(\mathbf{x}), f_2(\mathbf{x}), \ldots, f_m(\mathbf{x})\}$ ($\max$ also works in this context since $[m]$ has finite cardinality).
The context of a finite family is an important distinction since we don't technically invoke the Axiom of Choice; see: Axiom of Choice for Finite Sets. Moreover, the proof from the ordering principle of the axiom of choice for finite sets suggests why we should have an explicit definition such as $\max\{f( \mathbf x),g(\mathbf x)\}:=\frac12\left[f(\mathbf x)+g(\mathbf x)+|f(\mathbf x)-g(\mathbf x)|\right]$ since $\bigcup_{\mathbf x \in \mathbb R^n} \{f(\mathbf x),g(\mathbf x)\} \subset \mathbb R$ is already a totally ordered set without the ordering principle.
Best Answer
Since $f$ is continuous at $g(a)$, our definition of continuity tells us that for all $\varepsilon > 0$ there is some $\delta_1$ such that $$|g(x) - g(a)| < \delta_1\implies|f(g(x))-f(g(a))|<\varepsilon.$$ Also, since $g$ is continuous at $a$, there is some $\delta$ such that $$|x-a|<\delta \implies |g(x)-g(a)|<\delta_1.$$ I've taken $\varepsilon =\delta_1$ here. Now this tells us that for all $\varepsilon > 0$ there is some $\delta > 0$ (and a $\delta_1 > 0$) such that $$|x-a| < \delta\implies|g(x)-g(a)|<\delta_1\implies|f(g(x)) - f(g(a))|<\varepsilon,$$ which is what we wanted to show.