Suppose $f: \mathbb{R} \to \mathbb{R}$ is continuous at a. Prove $f \circ f: \mathbb{R} \to \mathbb{R}$ is continuous at $f(a)$. I am having a bit of difficulty with this proof.
I know $|x-a| < \delta \implies |f(x) - f(a)| < \epsilon$.
I know $f$ is only continuous at a, so do I need to prove f is continuous at $f(a)$ before moving on with the proof? If yes, I'm not sure how.
I have the first line $|f(x) - f(f(a))|$, but I'm not sure how to prove, in a general sense, how this is continuous.
Even bypassing proving the f(f(a)) continuity, the bulk of the problem is still an issue: Proving $|x-a| \implies f(f(x)) - f(f(a)) < \epsilon$.
Any help would be appreciated! 
 A: The statement doesn't hold always...
See the example $f:\mathbb{R} \rightarrow \mathbb{R}$:
$$ f(x)=\begin{cases}  x+2 &\text { when }x\in (-\infty,4) \\ 10 &\text {  when  }x\in [4,\infty] \end{cases} $$
Then: 
$$ (f\circ f)(x) = \begin{cases} x+4 &\text { when  }x\in (-\infty,2) \\ 10 &\text{ when } x\in [2,\infty]\end{cases} $$
Clearly $f $ is continuous at $2$, but $f\circ f$ is not continuous at $2$.
A: Some notes to help the OP sort out what the real problem is (to be deleted once the Question is appropriately clarified).
If $f$ were only defined at $x=a$, it would not make much sense to speak of it being continuous.  The definition of continuity at $x=a$ calls for $f$ to be defined on a deleted neighborhood of $a$.
Furthermore if $f(f(a))$ is defined, then $f(a)$ has to belong to the domain of $f$.  Unless we knew $f$ was continuous at $f(a)$ as well as at $x=a$, we would have no reason to expect $f\circ f$ to be continuous at $x=a$.  Counterexamples are easily constructed.
Possibly $f\cdot f$ is the product of $f$ with itself.  In this case, as @LL3.14 suggests, $f$ could be defined everywhere and $f\cdot f$ would be defined everywhere (on the real numbers), and continuity of $f$ at $x=a$ would imply continuity of $f\cdot f$ at $x=a$.
