Show that the comb space has fixed-point property By "comb space", I mean the space $X=([0,1] \times \{0\}) \cup (K \times [0,1])$, where $K=\{ \frac{1}{n} : n \in \mathbb{N}^+ \}$, without the leftmost vertical line segment.
How to prove that this space has the fixed-point property?
I've tried to embed $X$ as a retract of some space with fixed-point property, but this seems impossible since it is not a closed subspace of $\mathbb{R}^2$.

Let $f:X\to X$ be an arbitrary continuous map. We shall prove that $f$ has a fixed point.
Let $I=[0,1]\times\{0\}$ be the bottom interval and let $\pi:X\to I$ be the projection $\pi(x,y)=(x,0)$. The map $\pi\circ f$ maps $I$ to $I$ so it has a fixed point in $I$. Therefore, there exist $x_0$ and $y_0$ such that $$f(x_0,0)=(x_0,y_0).$$ Let $y_1$ be the largest number such that $f(\{x_0\}\times[0,y_1])\subseteq\{x_0\}\times[0,1]$. If $y_1=0$, we have $f(x_0,0)=(x_0,0)$ and we are done. If $y_1=1$, $f$ maps $\{x_0\}\times[0,1]$ to itself, and we again have a fixed point. So assume $0<y_1<1$. Let $$r(x_0,y)=(x_0,\min\{y,y_1\})$$ be the obvious retraction of $\{x_0\}\times[0,1]$ to $\{x_0\}\times[0,y_1]$. Then, $r\circ f$ maps $\{x_0\}\times[0,y_1]$ to itself, so it has a fixed point $(x_0,y)$ in $\{x_0\}\times[0,y_1]$. If this fixed point has $y<y_1$, we are done. Otherwise, $r(f(x_0,y_1))=(x_0,y_1)$, so $f(x_0,y_1)\in\{x_0\}\times[y_1,1]$. But then, for some $\epsilon>0$, we must have $f(\{x_0\}\times(y_1-\epsilon,y_1+\epsilon))\in\{x_0\}\times[0,1]$, contradicting the maximality of $y_1$. This completes the proof.
(If you need some more details, feel free to ask.)
