Hatcher's example of a pair without the HEP At the bottom of page 15, Hatcher gives the pair $(I,A)$ where $A= \left\{1,0,\frac 12,\frac 13,\dots \right\} $ as an example of a pair without the HEP. He then attributes the problem to the failure to the "bad structure of $(X,A)$ near $0$".
To disprove the HEP one need only show there is no continuous retraction $I\times I\rightarrow I\times \left\{ 0 \right\} \cup A\times I$. However, since retractions are surjective, it seems the disproof only uses the fact continuous images of connected spaces are connected along with the disconnectedness of $A$.  In particular, nothing about the "structure of $(X,A)$ near $0$".
What exactly does Hatcher mean by his statement?
Correction. As Mike Miller points out in the comments, $I\times \left\{ 0 \right\} \cup A\times I$ is connected. I mistakenly thought about $A$.
 A: It has something to do the failure of $A$ being locally connected near $0$. Roughly speaking, this means there are arbitrarily small neighborhoods of $0$ that are not connected, as can be verified directly.
Suppose $r$ is a retraction from $I\times I$ to $(A\times I)\cup (I\times 0)$. Let $p=(0, 1/2)$. We will show a contradiction by mapping, using $r$, a connected neighborhood of $p$ in $I\times I$ to a disconnected neighborhood of $p$ in $(A\times I)\cup (I\times 0)$. That disconnected neighborhood would simply be $U=A\times (0,1)$, whose components are $a\times (0,1)$, where $a\in A$; the connected neighborhood would be a half ball around $p$.
Since $r$ is continuous, we can find a half ball $B$ centered at $p$ such that $r(B)\subset U$. Since $B$ is connected, so is $r(B)$, which therefore should be contained in a single component of $U$. Since $p=(0,1/2)\in U$, that component must be $C=0\times (0,1)$. This means all of $B$ is mapped to $C$ under $r$. On the other hand, however, since $B$ is a neighborhood of $p$, there is some $n>0$ such that $(1/n,1/2)\in B$. Since $1/n\in A$, $r(1/n,1/2)=(1/n,1/2)\notin C$. This contradicts the previous claim that all of $r(B)$ is contained in $C$.
