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Can someone give me an example of the following:

Let $X$ be a topological space and $x_1, x_2 \in X$. Suppose that for every separation $X= A \cup B$, the points $x_1, x_2$ are both in $A$ or both in $B$. Give an example such that $x_1$ and $x_2$ are not in the same (connected) component of X.

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    Maybe this: $X=\{(0,0),(1,0)\}\cup \bigcup_{n \neq 1}\{(x, 1/n) \mid x\in \mathbb{R}\}$? And then take $x_1=(0,0)$ and $x_2 =(1,0)$. And then $x_1$ and $x_2$ are both either in $A$ or in $B$ because $A$ and $B$ should be open in $X$ and because these single points are not open in $X$ (limit points). Does that work? Or any other ideas? – 2017-01-26

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Put $X=\{ (0,0),(1,0)\cup \bigcup_{n\neq 1}\{ (u,1/n)|u\in\mathbb{R}\}\},\ x=(0,0),\ y=(1,0)$.

If there are two disjoint open not empty $A_{x},A_{y}\subseteq X$ such that $A_{x}\cup A_{y}=X$ then we can take, without lost of generality, $z\in\{(u,1/m):z\in\mathbb{R}\}\cap A_{x}$, for some $m\in\mathbb{N}$. Then $\{(u,1/m):z\in\mathbb{R}\}\cap A_{y}=\emptyset$, because if not, $\{(u,1/m):z\in\mathbb{R}\}\cap A_{x}$ and $\{(u,1/m):z\in\mathbb{R}\}\cap A_{y}$ is "disconnection" of $\{(u,1/m):z\in\mathbb{R}\}$. But $\{y\}$ is not open in $X$. $\rightarrow\leftarrow$

Since every $\{ (x,1/n)|x\in\mathbb{R}\}$ is a connected component of $X$, we deduce that $x$ and $y$ are in the same component.