Suppose f is a continous map from $(X,d)$ to $(Y,p)$. Suppose A is a subspace of $X$. Then the restriction of f from $A$ to Y is likewise continous.
My proof:
Let $a \in A$. Suppose $(a_n)$ is an arbitrary sequence contained in $A$ such that $a_n \rightarrow a$. Since $f$ is continous on X, $f(a_n) \rightarrow f(a)$.
Is this proof correct? How could I make it more detailed? What else should I include? Could you provide another proof?
For the sake of providing an answer rather than a comment:
Looks entirely correct. Other proofs would depend on your definition of continuity. Being a topologist, my definition of continuity is that "pre-images of open sets are open". In your context these two definitions are equivalent, so you could try proving that all pre-images of open sets are open?
A sketch of my alternative proof would be something like:
$f:X \rightarrow Y$ is continuous, so for any open set $B \subseteq Y$, its preimage $f^{-1}(B)$ is open.
I now look at the open sets in the subspace $A \subseteq X$. There is possibly some lemma that would help me here.
I now deduce that if $f^{-1}(B)$ is open in $X$, then ${\widehat{f}}^{-1}(B)$ is open in $A$, where $\widehat{f}$ is the restriction of $f$ to $A$.
That should give you the bare bones.