subspaces of $\mathbb{R}^2$ that are not retracts of $\mathbb{R}^2$. I have to either prove or disprove: There exist infinitely many subspaces (up to homeomorphism) of $\mathbb{R}^2$ that are not retracts of $\mathbb{R}^2$.
I can think of a few subspaces (e.g., finite sets) that are not retracts. A hint or two to get me going in the right direction will be greatly appreciated!
W.
 A: Consider $S^1\times \{1,2,\ldots, n\}\subseteq\Bbb R^2$ embedded by looking at the collections of circles of radius $1$ centered around say the first $n$ even natural numbers, i.e.
$$\bigcup_{k=1}^n\left\{(x,y)\in\Bbb R^2: (x-2k)^2+(y-2k)^2=1\right\}$$ There is one for each $n\in\Bbb N$. Clearly they are not homeomorphic just by counting connected components, and none are retracts of $\Bbb R^2$, which is observable by checking the fundamental groups, which are $\Bbb Z$ (choose any base-point).
Your example of finite sets can work too, clearly
$$[n]=\{1,2,\ldots, n\}$$
are all non-homeomorphic by counting connected components, but since the fundamental groups are trivial, it takes a bit more set theory to push things around to work, and you only get it for $n>1$, obviously, but that's not terrible either. Mostly you just say $\Bbb R^2\to [n]$ is necessarily constant because $\Bbb R^2$ is connected, then since retractions have left-inverses, by definition, you have
$$[n]\to \Bbb R^2\to [n]$$
is the identity, but this only works if $n=1$.
