Every retraction is a quotient map?

You could use the following theorem; Let $f: X \rightarrow Y$ be a continous map and suppose there is a continuous map $g : Y \rightarrow X$ such that $f \circ g$ is the identity. Then $f$ is a quotient map.

Now let $r:X \rightarrow A$ be your retraction, you could now easiy find a map $s$ such that $r \circ s$ is the identity map on A.

HINT: By definition the map $r$ is a quotient map if and only if the following is true:

$\qquad\qquad\qquad$ a set $U\subseteq A$ is open in $A$ if and only if $r^{-1}[U]$ is open in $X$.

Suppose that $U\subseteq A$ is open in $A$; then $r^{-1}[U]$ is certainly open in $X$, simply because $r$ is continuous. Now suppose that $U\subseteq A$ is not open in $A$, and let $S=r^{-1}[U]$; you want to prove that $S$ is not open in $X$. If $S$ were open in $X$, then $S\cap A$ would be open in $A$, by definition of the subspace topology on $A$. What is $S\cap A$? Is it open in $A$?

If you don't want to play around with open sets, you could use the universal property of quotient maps: a continuous map $q:X\to A$ is a quotient map if and only if, whenever $f:X\to Z$ is a continuous map such that $q(x)=q(y)\Rightarrow f(y)=f(y)$, there is a unique continuous map $f':A\to Z$ such that $f = f'\circ q$. I suggest that in the case where $q$ is a retraction onto $A\subset X$, given such an $f$, there is a clear choice of $f'$ to try.