Infinite countable subspace of $\mathbb{R}^2$ is not connected

Question

I've seen different proofs of this fact, but none of them look like mine which is kind of simple; so I guess I am wrong but I cannot find the flaw in the following line of reasoning:

Let $A$ be an infinite countable subspace of $\mathbb{R}^2$. Assume, by way of contradiction, $A$ is connected. Then $\pi_x(A)$ is a connected subset of $\mathbb{R}$ as $\pi_x$ (the projection onto the first coordinate) is a continuous function. Since the connected subspaces of $\mathbb{R}$ are intervals, either $\pi_x$ is an interval with at least two elements or $\pi_x$ is a singleton. In the first case, the contradiction is immediate; as for the second case we can repeat this procedure projecting onto the second coordinate.

Can you please point out my mistake? Thanks.

Answer 1

This does work, although I'd phrase it a bit differently:

Suppose $A\subseteq\mathbb{R}^2$ is connected and has at least two distinct elements. Now consider the projections $\pi_1(A)$ and $\pi_2(A)$ of $A$ onto the $x$ and $y$ axes, respectively.

Since $A$ has at least two distinct elements, we can't have both $\pi_1(A)$ and $\pi_2(A)$ be singletons (since if $(x,y)\not=(u,v)$ either $x\not=u$ or $y\not=v$). Suppose WLOG that $\pi_1(A)$ has more than one element. Since $\pi_1$ is continuous and $A$ is connected, $\pi_1(A)$ is a connected subset of $\mathbb{R}$ with at least two elements. But then $\pi_1(A)$ must contain a nontrivial open interval, hence be uncountable.

Answer 2

Your proof is correct. It shows that the projection of $A$ onto each coordinate can only be a singleton, which implies $A$ itself must be a singleton. The same argument works for subsets of $\mathbb{R}^n$ for any $n$ (in fact, even for infinite products $\mathbb{R}^I$ for arbitrary sets $I$).