# Comb space is contractible but not base point preserving

For each positive integer $n$, let $I_n=\{1/n\}\times I$ as a subset of $I\times I$. Let $X=(I\times0)\cup(0\times I)\cup(\bigcup_{n\geq 1} I_n).$ Let $x_0=(0,1)\in X$ be the base point. Show that $X$ is contractible. Show, however, that there is no base point preserving homotopy between the identity map of $X$ and the constant map at $x_0$.

I know that it is contractible. Is there any hint about the latter?

Idea: Take the sequence $x_n=(\frac{1}{n},1)$, it converges to $x_0$. If existed such homotopy $H(x,t)$ then the sequences $H(x_{n},t)$ would still converge to $x_{0}$.
You have for each neighborhood $U$ of $x_0$ a number $N_{(U,t)}$ and $\epsilon_{(U,t)}$ such that $H(x_n,s)\in U$ for $n>N_{(U,t)}$ and $|t-s|<\epsilon_{(U,t)}$.
Covering the interval $I$ with $(t-\epsilon_{(U,t)},t+\epsilon_{(U,t)})$ and taking a finite subcover you obtain a number $N$ such that $H(x_n,t)\in U$ for any $n>N$ and $t\in[0,1]$.
Now you can use disconnectedness of a small neighborhood to show that the homotopy can't take the elements of the sequence to $x_0$.