Topology: Determine whether a subset is a retract of R^2 1. The problem statement, all variables and given/known data
Let $X=([1,\infty)\times\{0\})\cup(\cup_{n=1}^{\infty}\{n\}\times[0,1])$ and $Y=((0,\infty)\times\{0\})\cup(\cup_{n=1}^{\infty}\{n\}\times[0,1])$
$a)$Find   subspaces of of the euclidean plane $\mathbb{R}^2$ which are homeomorphic to the compactification with one point $X^+$ and $Y^+$
$b)$ are any of the subspaces $X$ and $Y$ retracts of $\mathbb{R}^2$
2. Relevant equations
3. The attempt at a solution
I was able to find the required subspaces for $a)$ and proved that they are homeomorphic, however I'm having some problems with $b)$. I know that $Y$ cannot be a retract as it is not closed.
I got that since $\mathbb{R}^2$ is a metric space  therefore $T_{2}$ and we know that if $f,g:X\to Y$ are continuous and $Y\in T_{2}$ then the set $\{x|f(x)=g(x)\}\subset Y$ is closed. However in this case $Y=\mathbb{R}^2$ and if there exists a retraction $r:\mathbb{R}^2\to Y\subset \mathbb{R^2} $ the set $\{x|r(x)=id_{x}(x)=x\}$,which is  $X$, must be closed. And since it is not.$ X $ cannot be a retraction. Can someone correct me here if I did something wrong?
However this argument does not work for $X$ as $X$ is closed. $X$ is also connected so that fails too. I can't find a homeomorphism between $X$ and $\mathbb{R}$. Could someone clarify how we could determine whether X is a retract or not?
 A: You're correct about space $Y$ not being a retract of $\mathbb R^2$ due to it being non-closed. 
For the second part, it's enough to show that $X$ is a retract of $A:=[1,\infty)\times[0,1]$ since $A$ is a retract of $\mathbb R^2$, because there holds that if $X$ is a retract of $A$ and $A$ is a retract of $\mathbb R^2$, then $X$ is a retract of $\mathbb R^2$ (this "transitivity" follows straightforwardly from the definition of a retract).
Let $X_n:=\{n\}\times[0,1]\cup[n,n+1]\times\{0\}$ and $A_n:=[n,n+1]\times[0,1]$. Let's first prove $X_n$ is a retract of $A_n$. $X_n\cong [0,1]$, i.e. there exists a homeomorphism $f_n\colon X_n\rightarrow [0,1]$. Then by Tietze's extension theorem, $\exists \hat f_n:A_n\rightarrow[0,1]$ continuous, such that $\hat f_n|_{X_n}\equiv f_n$. Then $r_n\colon A_n\rightarrow X_n$ given by $r_n=f_n^{-1}\circ \hat f_n$ is a retraction of $A_n$ to $X_n$. Now define $r\colon \cup_{n=1}^\infty A_n\rightarrow\cup_{n=1}^\infty X_n$ piecewise, that is, $r(x,y)=r_n(x,y)$ whenever $(x,y)\in A_n$. $r$ is continuous, because it is piecewise continuous on a locally finite closed cover $\{A_n\;|\;n\in\mathbb N\}$ of $A$ and there holds $r|_X\equiv \mathrm{id}_X$ since the restriction is the identity piecewise. Therefore $X$ is a retract of $A$.
Edit: maybe I was too hasty to say that $A$ is a retract of $\mathbb R^2$, so here's proof. Identity map $\mathrm{id}_A\colon A\rightarrow A$ is continuous and $A\in\operatorname{AE}(\mathcal N)$, because $A$ is a product of intervals $[1,\infty)$, $[0,1]$ from $\operatorname{AE}(\mathcal N)$ (here $\mathcal N$ denotes the class of normal spaces). Therefore, $\exists\tilde r\colon \mathbb R^2\rightarrow A$, such that $\tilde r|_A\equiv \mathrm {id}_A$. Finally, $r\circ \tilde r\colon \mathbb R^2\rightarrow X$ is our wanted retraction.
