Show that the Moore plane is not normal 
Definition: A Hausdorff space is normal (or: $T_4$) if each pair of disjoint closed sets have disjoint neighborhoods.

Then, we have

Exercise 5, pg. 158, Dugundji's Topology: Let $X$ be the upper of the Euclidean plane $E^2$, bounded by the $x$-axis. Use the Euclidean topology on $\{(x,y)\,|\, y>0\}$, but define neighborhoods of the points $(x,0)$ to be $\{(x,0)\}\cup [\text{open disc in $\{(x,y)\,|\,y>0\}$ tangent to the $x$-axis at $(x,0)$}]$. Prove that this space is not normal.

It is easy to see that $X$ is Hausdorff. So, what we need is to find a pair of closed sets that fail to satisfy the definition given above. Here, Alice Munro says that $A=\{(x,0)\,|\, x\in \Bbb Q\}$ and $B=\{(x,0)\,|\, x\in \Bbb R-\Bbb Q\}$ are such sets. I can see that they are closed, but how can I show that they do not admit disjoint neighborhoods? (intuitively true, but I'm having difficulty to write it down...)

My preferred way (though the statement that these two sets are closed disjoint sets that cannot be separated is true) is to use a well-known cardinal number fact in normal spaces often called Jones' lemma:

Let $X$ be $T_4$. If $D$ is a dense subset of $X$ and $C$ is a closed and discrete (as a subspace) subset of $X$, then $2^{|C|} \le 2^{|D|}$.

I prove it in this answer; it uses Urysohn's extension theorem for normal spaces: we need a lot of different continuous real-valued functions on $X$ to separate all subsets of $C$, and $D$ bounds that number.
There I also show how it applies to the Moore plane (or Niemytzki plane): the $x$-axis is closed and discrete and $\mathbb{Q} \times \mathbb{Q}^+$ is countable and dense and $2^{|\mathbb{R}|} \not\le 2^{\aleph_0} = |\mathbb{R}|$.
