Real Analysis: Continuity of a Composition Function

Suppose $f$ and $g$ are functions such that $g$ is continuous at $a$, and $f$ is continuous at $g(a)$. Show the composition $f(g(x))$ is continuous at $a$.

My idea: Can I go straight from definition and take $\delta=\min\{\delta_1,\delta_2\}$, where $\delta_1$ is used for the continuity of $g$ at $a$ and $\delta_2$ is used for f being continuous at $g(a)$. In my proof I just treat $g(a)$ as a point when referring to the composition. So it goes like this:

Proof: Given $\epsilon>0$, take $\delta=\min\{\delta_1,\delta_2\}$. Then $0<|x-g(a)|<\delta$ which implies $|f(g(x))-f(g(a))|<\epsilon$.

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.

Proof

It will be shown that the limit of $f(g(x))$ at any arbitrary point $x=a$ in the domain of $f(g(x))$ is equal to $f(g(a))$.

1. Let $a_n$ be any convergent sequence such that $a_n\to a$.
2. Since $g(x)$ is continuous, $g(a_n)\to g(a)$ as $a_n\to a$.
3. Since $f(x)$ is continuous, $f(g(a_n))\to f(g(a))$ as $a_n\to a$ as required.

If $f \circ g$ is well defined then without any loss of generality let's assume that $f:A\to R $, and $g:B \to R$ such that $f(A) \subset B$. Also $A$ and $B$ are $\subset R$.

Now since $g$ is continuous at $a$, given $\epsilon >0$,there exists a $δ >0$, such that whenever $0<|x−a|<δ$, we have $|f(x)−f(a)|<ϵ$. Now choose $\delta_1 >0$ such that whenever $0<|y−a|<δ_1$ $\implies |g(y)−g(a)|< \delta$.

Then we are done.