The other day, I was idly considering when a topological space has a square root. That is, what spaces are homeomorphic to $X \times X$ for some space $X$. $\mathbb{R}$ is not such a space: If $X \times X$ were homeomorphic to $\mathbb{R}$, then $X$ would be path connected. But then $X \times X$ minus a point would also be path connected. But $\mathbb{R}$ minus a point is not path connected.

A next natural space to consider is $\mathbb{R}^3$. My intuition is that $\mathbb{R}^3$ also doesn't have a square root. And I'm guessing there's a nice algebraic topology proof. But that's not technology I'm much practiced with. And I don't trust my intuition too much for questions like this.

So, is there a space $X$ so that $X \times X$ is homeomorphic to $\mathbb{R}^3$?

No such space exists.  Even better, let's generalize your proof by converting information about path components into homology groups.

For an open inclusion of spaces $X \setminus \{x\} \subset X$ and a field $k$, we have isomorphisms (the relative Kunneth formula)
$$
H_n(X \times X, X \times X \setminus \{(x,x)\}; k) \cong \bigoplus_{p+q=n} H_p(X,X \setminus \{x\};k) \otimes_k H_q(X, X \setminus \{x\};k).
$$
If the product is $\mathbb{R}^3$, then the left-hand side is $k$ in degree 3 and zero otherwise, so something on the right-hand side must be nontrivial.  However, if $H_p(X, X \setminus \{x\};k)$ were nontrivial in degree $n$, then the left-hand side must be nontrivial in degree $2n$.
