Restriction of Continuous Functions is Continuous (Epsilon-Delta) Most of the proofs given to verify that the restriction of continouous functions is continuous, require the use of topological concepts. I was wondering if it were possible to do so using the $\epsilon-\delta$ definition of continuity.
From Introduction to Real Analysis: 
$f:A\rightarrow R $
$f$  is continuous at $a\in A$ iff
$$\forall \epsilon>0\ \exists \delta>0\ (\forall x\in(a-\delta,a+\delta)\cap A \implies f(x)\in(f(a)-\epsilon,f(a)+\epsilon)).$$
Suppose we restrict $f|_Q:Q \rightarrow R$ where $Q \subseteq A$.
For the restriction, wouldn't the same $\delta$ used in the original function work?
 A: Yes, the same $\delta$ works. Let $g:=f|_Q:Q\to\mathbb{R}$ and $\varepsilon>0$. Since $f$ is continuous, for each $a\in Q$ there exists $\delta>0$ such that whenever $x\in A$, then
$$
|x-a|<\delta \implies |f(x)-f(a)|<\varepsilon.
$$
Hence, if $x\in Q\subseteq A$, then
$$
|x-a|<\delta \implies |g(x)-g(a)|=|f(x)-f(a)|<\varepsilon.
$$

Edit: Comment about proof versus the "topological one".
This result for metric spaces is part of the motivation for the subspace topology to be defined in the way that it is. Indeed, when defining the subspace topology on a subset $A$ of a topological space $X$, we want to do it in such a way that the inclusion map from $A$ into $X$ (which is just the restriction of the identity map on $X$) is continuous. Thus the moral of the story becomes:
If we are dealing with metric spaces, this is the "correct" proof because it doesn't require showing that metric space continuity is consistent with that given by the topological definitions. If we are dealing with general topological spaces, then the definitions are cooked up precisely so that this is trivial.
