Homeomorphism between two subspaces of $\mathbb{R}^n$ Let $$X_1 = \left\{(0,0)\right\} \cup \left\{\left(0,\frac{1}{n}\right): n = 1,2... \right\} \cup \left\{\left(1,\frac{1}{n}\right): n = 1,2... \right\}$$
$$X_2=\left\{(0,0)\right\} \cup \left\{\left(\frac1n,\frac1m\right): n \ge m, m,n\in\mathbb{N}\right\}$$
Decide whether those sets are homeomorhic as subsets of $(\mathbb{R}^2, d_e)$.
I suspect they are not, but I cannot find a good reason why. They both have one limit point, so the easiest argument doesn't work. Also in the neighbourhood of this point and cannot see anything special.
Edit. Is it true that neighbourhoods of limit points are not homeomorphic, because one is compact and the other is not? If so, is it enough?

The space $X_1$ can be decomposed into two clopen sets (closed-and-open), $$X_1=C_1 \cup C_2
$$
where $C_1$ is a compact set containing the unique limit point, and $C_2$  is a countably infinite discrete set. Namely, $C_1=\left\{(0,0)\right\} \cup \left\{\left(0,\frac{1}{n}\right): n = 1,2... \right\}$ and $C_2=\left\{\left(1,\frac{1}{n}\right): n = 1,2... \right\}$.
But $X_2$ has no such decomposition, hence $X_1$, $X_2$ are not homeomorphic. 
To see why, suppose that $X_2$ does have such a decomposition, 
$$X_2 = C_1 \cup C_2
$$
The set $C_2$ is disjoint from some neighborhood of the unique limit point $(0,0)$ of $X_2$. Since $C_1$ is open, it follows that $C_2 \subset \{(x,y) : y \ge \epsilon\}$ for some $\epsilon>0$. The set $C_1$ therefore contains a sequence with no convergent subsequence, namely $(1/n,1/m)$ for some fixed value of $m > 1/\epsilon$. That contradicts compactness of $C_1$.
