Retractions and local finiteness: Are these circles a retract of $\mathbb R^2$? Let $N$=$\{1+\frac 1 n\;|\;n\in\mathbb N\}\cup\{1\}$. I'm interested in whether the space
$$
A=\cup_{n\in N}A_n \text{ where } A_n:=\{(x,y)\in\mathbb R^2\;|\;(x-n)^2 + y^2 = n^2, x\leq 1\}
$$
is a retract of $\mathbb R^2$. Note that since $1\in N$, the limiting circle is also in $A$. The space is the red part of the following picture:

First, observe that the space is path connected, since all circles intersect in $(0,0)$. Secondly, the space is closed, because it is an intersection of a descending chain of closed sets (or, if you prefer, because all of its accumulation points are its elements). Since we fail to conclude with both of these arguments that $A$ is not a retract of $\mathbb R^2$, we next try to prove that it is a retract. 
I've tried these things, none of which worked out:


*

*My first idea was to try and combine "croissants" (you'll see what I mean if you look at the picture) into the disk $\mathbb D((2,0),2)$. It's possible to show that for $n\geq 2$ $A_n\cup A_{n+1}$ is a retract of the croissant $C_n=\{(x,y)\in\mathbb R^2\;|\;(x-n)^2 + y^2 \leq n^2\text{ and }(x-n\frac{n-1}{n+1})^2 + y^2 \geq (n\frac{n-1}{n+1})^2\}$
and also that $A_1$ is a retract of $\mathbb D((1,0), 1)$. (I won't go into detail of how to do it as it is not so relevant.) When I tried to combine the retractions piecewisely on the closed cover $\{C_n\;|\;n\in N\setminus\{1\}\}\cup \{\mathbb D((1,0),1)\}$ of $\mathbb D((2,0),2)$ into a continuous map $r\colon\mathbb D((2,0),2)\rightarrow A$, I encountered two problems. Firstly, there was no reason to think that the retraction maps $r_n\colon C_n\rightarrow A_n\cup A_{n+1}$ are consistent on their intersections, so the combined map isn't even well-defined. Secondly, $(0,0)$ is problematic; since there is no neighborhood of $(0,0)$ that would intersect only finitely many sets in the closed cover (i.e. the given cover is not locally finite), we cannot guarantee that the piecewisely defined map $r$ is continuous. This second issue essentially means that if you try to come up with a piecewisely defined retraction of some subspace of $\mathbb R^2$ to $A$, you will have problems with continuity, because the point $(0,0)$ doesn't have a neighborhood that would intersect finitely many $A_n$'s. I also tried to combine retractions of the circles $S_n=\{(x,y)\in\mathbb R^2\;|\;(x-n)^2 + y^2 = n^2\}$ to $A_n$ to yield a retraction of $\cup_{n\in N}S_n$ to A (if that could, in a separate attempt, somehow help us to get a retraction of $\mathbb R^2$ to $A$) and encountered the same issue.

*So, I began to suspect that $A$ is not a retract of $\mathbb R^2$. Since we know that every retract of an absolute extensor is itself an absolute extensor, we could somehow use it to prove with a contradiction that $A$ is not a retract of $\mathbb R^2$ (e.g. proving that extension from some subspace to the whole space of some continuous map to $A$ isn't continuous). Also, $A$ is compact, which could also be something that we could use. So, assume that $A$ is a retract of $\mathbb R^2$. Then, $A$ is also an absolute extensor of normal spaces. $\mathrm{id}_A$ is a continuous map and therefore there exists a function $r\colon \mathbb R^2\rightarrow A$ such that $r|_A\equiv \mathrm{id}_A$. But I have not a clue what to do from here on.


At some point in yielding a contradiction, we should probably use that some chosen cover isn't locally finite, but I don't really know where that will help me.
 A: $\mathbb{R}^2$ is an Absolute Retract by Dugundji.
Assume that your $A$ is a retract of $\mathbb{R}^2$. Then $A$ is an AR as well. And since a space is an AR if and only if it is contractible and an Absolute Neighbourhood Retract (Dugundji as well?), then $A$ is an ANR.
Every ANR is locally contractible and thus locally connected but $A$ is not locally connected. Contradiction.
A: The question was posted after an exam in my general topology class. I've now had my first geometric topology class and came back to answer this.
I have constructed a solution to this problem that is way more elementary than that of @freakish. Assume that $r\colon \mathbb R^2\rightarrow A$ is a retraction. Let 
$$
I_n=\{1\}\times [y_{m_n},y_{m_{n+1}}] \quad \forall n\in\mathbb N 
$$
where $y_m=\sqrt{m^2-(1-m)^2}$ and $m_n=1+\frac 1 n$. Let $n\in\mathbb N$. $I_n$ is a line that connects the endpoints of red arcs in the picture in the original post. Because $I_n$ is connected, $r(I_n)$ is too. Since $r(1,y_{m_n})=(1,y_{m_n})$ and also $r(1,y_{m_{n+1}})=(1,y_{m_{n+1}})$, there must hold 
$$
(A_n\cup A_{n+1})\cap([0,\infty)\times[0,\infty))\subset r(I_n),
$$
thus $\exists a_n\in I_n$ such that $r(a_n)=(0,0)$. Now 
$$
\lim_{n\rightarrow\infty} a_n = (1,1)
$$
but 
$$
(1,1)=r\left(\lim_{n\rightarrow\infty}a_n\right)=\lim_{n\rightarrow\infty}r(a_n)=(0,0).
$$
Contradiction.
