Problem Statement:
Let $X = (\bigcup \limits_{n \in \mathbb N} \{\frac{1}{n}\} \times [0,1] ) \cup \{(0,0),(0,1)\}$ have a subspace topology as a subspace of $\mathbb R^2$. For any separation $U$ and $V$ of $X$, if $(0, 0) \in U$, then $(0, 1) \in U$ as well.
My attempt:
By the result from Munkres, if $U$ and $V$ are a separation of $X$ and $Y$ is a connected subspace of $X$, then $Y$ is completely contained in either $U$ or $V$. Hence, to show that $(0, 0) \in U$ would imply $(0, 1) \in U$, it suffices to show that there exists some connected subspace of $X$ that contains both $(0, 0)$ and $(0, 1)$.
From here, I am having trouble finding some connected subspace of $X$ that contains both points.
Best Answer
Your solution will not work. Any neighbourhood $W$ of $X$ which contains both $(0,1)$ and $(0,0)$ can be disconnected either by the open sets $\{(x,y)|y>\frac12\}$ and $\{(x,y)|y<\frac12\}$ or if there exists $n\in\mathbb{N}$ with $(\frac1n,\frac12)\in W$, by the open sets $\{(x,y)|x<\frac1n\}$ and $\{(x,y)|x>\frac1{n+1}\}$.
Instead you can use the following argument:
If $U$ and $V$ separate $X$, then each vertical line $L_n=\{\frac1n\}\times [0,1]$ is connected, so by the result you mentioned completely contained in $U$ or $V$.
Any neighbourhood $U$ of $(0,0)$ will intersect all vertical lines $L_n$ for all $n>m_1$ for some $m_1$.
Similarly, any neighbourhood $V$ of $(0,1)$ will intersect all vertical lines $L_n$ for all $n>m_m$ for some $m_2$.
Thus if $U,V$ separate $X$, they will each contain all $L_n$ for $n>\max(m_1,m_2)$, yielding the desired contradiction.