I'm having difficulty seeing how $X=\{x\times y:y=0\}$ is open in $\mathbb{R}^2$. Being open would imply for any point $x\in X$, I can find a neighborhood around it contained in $X$, but I don't think any such neighborhood $(a,b)\times(c,d)$ is contained in $X$, as taking product with $c\times d$ takes us out of $X$.
The topology of $\mathbb{R}$ is the usual topology.
I ask this question, because in Munkres Section 23, Example 5, the author states $X=\{x\times y:y=0 \}\cup \{x\times y:x>0\land y=1/x\}$ is not connected… And, I just realize as I'm typing this that in the subspace topology, a separation need not consist of open sets.
Thanks to everyone that replied.
Best Answer
Let $X$ be the set in Munkre's exercise. $\{(x,y)\in \mathbb R^{2}: xy<1\}$ and $\{(x,y)\in \mathbb R^{2}: xy>\frac 1 2\}$ are open sets in $\mathbb R^{2}$. Take their intersections with $X$ to get two disjoint open sets in the subspace topology whose union is $X$.