If $f_1, \ldots, f_n : \Bbb{R} \to \Bbb{R}$ are continuous then $f_n \circ f_{n-1} \circ \ldots \circ f_1$ is continuous. Proof by induction. Proof by induction.
Induction basis: $n=1$. $f_1=f_1$.
Let an be any convergent sequence such that $a_n \to a$. Since $f_{n-1}(x)$ is continuous, $f_{n-1}(a_n) \to f_{n-1}(a)$ as $a_n \to a$. Since $f_n(x)$ is continuous, $f_n(f_{n-1}(a_n)) \to f_n(f_{n-1}(a))$ as $a_n \to a$ as required.
I'm not sure if this is sufficient to prove the case. Do I need to get into an induction proof using $\varepsilon$ and $\delta$?
 A: You don't need to get into $\varepsilon$s and $\delta$s; sequential continuity is fine.
That said, you've only established that $f_n \circ f_{n-1}$ is continuous, not $f_n \circ f_{n-1} \circ \ldots \circ f_1$. You also haven't used the assumption that $f_{n-1} \circ f_{n-2} \circ \ldots \circ f_1$ is continuous, which is your inductive hypothesis. Basically, your proof can be fixed up by replacing $f_{n-1}$ with $f_{n-1} \circ \ldots \circ f_1$ in your inductive step.
That is, if $a_m \to a$, then by assumption,
$$(f_{n-1} \circ \ldots \circ f_1)(a_m) \to (f_{n-1} \circ \ldots \circ f_1)(a)$$
as $m \to \infty$. Since $f_n$ is continuous,
$$f_n((f_{n-1} \circ \ldots \circ f_1)(a_m)) \to f_n((f_{n-1} \circ \ldots \circ f_1)(a)),$$
that is,
$$(f_n \circ f_{n-1} \circ \ldots \circ f_1)(a_m) \to (f_n \circ f_{n-1} \circ \ldots \circ f_1)(a)$$
as $m \to \infty$.
Note also the use of $m$ instead of $n$. This is important, because the inductive hypothesis fixes $n$. You assume the proposition holds for some $n$ that is fixed, but unknown. If you want to use a separate sequence like $(a_m)$, pick another pronumeral.
A: No, you don't need to use $\epsilon-\delta$ here, but note that you have assumed that $f_n \circ f_{n-1}$ is continuous, but you are actually required to prove it.
We assume the induction hypothesis till $n-1$ step and then show it also holds for $n$-th step. Though here it is easy as we have that $f_n$ is also continuous. So, if $a_n\rightarrow a$ we have $f_{n-1}\circ f_{n-2}...\circ f_1(a_n)\rightarrow f_{n-1}\circ f_{n-2}...\circ f_1(a)$ which is a  convergent sequene itself, so $f_n\circ (f_{n-1}\circ f_{n-2}...\circ f_1(a_n)) \rightarrow f_n\circ (f_{n-1}\circ f_{n-2}...\circ f_1(a)).$ And we prove the result. $\\$
Also, note that here we have used the fact that the composition of two continuous functions is continuous.
Hope this helps.
