Homotopy equivalence with a point need not be a deformation retract

This question arises from problem 8 on pg. 366 of Munkres. Let $X$ be the union of the sets $(1/n) \times I$, $0 \times I$, and $I \times 0$, where $I = [0,1]$, with the topology it inherits from $\mathbb R^2$. I am trying to prove that the inclusion map from the singleton set $x_0 = (0,1)$ to $X$ is a homotopy equivalence, but that $x_0$ is not a deformation retract of $X$.

Intuitively, my idea is as follows: in order to obtain a homotopy from the constant map $x \mapsto x_0$ to the identity map on $X$, we must contract the entire space $X$ down to the horizontal axis, then slide it back up the vertical axis to $x_0$. But, we cannot do so while keeping $x_0$ fixed because (in order to preserve continuity) the horizontal axis must slide down along with the lines $(1/n) \times I$, which are arbitrarily close to it. However, I am having a hard time formalizing this notion. Any suggestions?

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You may also be interested in looking at this (more complicated) variation of the comb space. – t.b. Dec 22 '11 at 17:49

Suppose $F: X \times I \to X$ is a retraction, and consider the sequence $x_n = (1/n, 1)$. Since $x_n \to x_0$, by continuity $F(x_n, t_0) \to F(x_0, t_0) = x_0$ for all $t_0$.
Now, take a small enough neighborhood $U$ of $x_0$. By the convergence above, there exist a neighborhood $V_{t_0}$ of $t_0$, and $n_0$ such that, for $t \in V$ and $n \geq n_0$, $F(x_n, t) \in U$. But $U$ can be chosen such that $X_{n} = (\{1/n\} \times I) \cap U$ is clopen in $U$ for all $n$, so $F(x_{n}, t) \in X_{n'}$ for all $t \in V_t$, $n \geq n_0$ and some $n'$.
By using compactness of $[0, 1]$, we can extract a finite subcover from all the $V_t$, and we obtain that for sufficiently large $n$, $F(x_n, t) \in X_{n'}$ for all $t$ including $0$ and $1$, contradiction.