If X is an absolute retract and Hausdorff, and A is a retract of X, then A is an absolute retract Let $X$ be a $T_2$ space and an absolute retract (AR). Further, let $A \subset X$ be a retract of $X$. Then $A$ is an AR.

In a $T_2$ space: for all distinct $x, y \in X$ there are open and disjoint $U,V$ containing $x$ and $y$ respectively.
For a retract $A$ of $X$: there exists a continuous mapping $r: X \mapsto A$ such that $r\vert_A=\mathrm{id}$.
For an AR $X$: whenever $X$ is embedded as a closed subspace of a normal space $Y$, then $X$ is a retract of $Y$.

I have tried to make use of these definitions, but have gotten nowhere. I must show that for any embedding (as a closed subspace) into a normal space, there is a retraction $r$. I struggle to find how normality and the aspect of a closed subspace play into this. Could you give me some hints as to how to proceed?
 A: Suppose that $f:A\hookrightarrow Y$ is an embedding of $A$ into a normal space $Y$. We must show that $A$ is a retract of $Y$.
Begin by forming the adjunction space $Z=X\cup_A Y$. Note that $X$ is canonically a closed subspace of $Z$.

Lemma: $Z$ is normal.

Proof: It is easy to see that $Z$ is $T_1$ when both $X$ and $Y$ are. Therefore it will suffice to show that whenever $C\subseteq Z$ is closed, any continuous $\alpha:C\rightarrow I$ extends over $Z$ (this is the Tietze Extension Theorem).
We start with the fact that $C\cap X$ is closed in the normal $X$ to find an extension $\beta:X\rightarrow I$ of the restriction $\alpha|_{C\cap X}$. From this we obtain a well-defined function $\beta|_A\cup \alpha|_{C\cap Y}:A\cup(C\cap Y)\rightarrow I$, which is continuous because $A$ and $C$ are closed, and hence extends to $\gamma:Y\rightarrow I$ by virtue of the normality of $Y$. Because the functions $\beta$ and $\gamma$ agree on $A$, the universal property of the adjunction space defines a continuous map $\widetilde\alpha:Z\rightarrow I$. This map satisfies $\widetilde\alpha|_C=\alpha$. $\square$
Returning to the initial problem we have the assumption that $X$ is an absolute retract for normal spaces. Therefore, by the lemma, there is a retraction $r:Z\rightarrow X$. Our second assumption is the presence of a retraction $s:X\rightarrow A$. Putting these maps together we have the composite
$$t:Y\hookrightarrow Z\xrightarrow{r}X\xrightarrow sA,$$
which is a retraction of $Y$ onto $A$.
