Continuum: $X$ is a continuum if it is a compact connected Hausdorff space.
A continuum $X$ cannot be equal to $\bigcup\limits_{n=1}^\infty F_n$ for any $\{F_n\}$, where $\{F_n\}$ is a collection of disjoint nonempty closed subsets of $X$.
Hint: Let $X = \bigcup\limits_{n=1}^\infty F_n$, and $U_1$ be open set containing $F_2$ such that $\overline U_1 \cap F_1 = \varnothing$.
Let $X_1$ be a component of $\overline U_1$ meeting $F_2$. Then $X_1 \cap F_1 = \varnothing$ , and $X_1$ meets some $F_{n_2}, n_2>2$.
Let $U_2$ be open set containing $F_{n_2}$ such that $\overline U_2 \cap F_2 = \varnothing$, and let $X_2$ be a component of $\overline U_2\cap X_1$ meeting $F_{n_2}$.
Continuing, show that $X_1\supseteq X_2\supseteq X_3\supseteq\cdots$, but $\bigcap X_n = \varnothing$, obtaining a contradiction.
I was able to show that the sequence $(X_n)$ is a decreasing sequence of continua. However, I haven't been able to show that $\bigcap X_n = \varnothing$. Any help is appreciated!
Best Answer
Note that in your construction,
$$ X_n \cap F_n = \varnothing.$$
Since $\bigcap_k X_k$ is in $X_n$ for all $n$, $\bigcap_k X_k \cap F_n = \varnothing$ for all $n$. Since we assume $X = \bigcup_n F_n$, we have $$ \bigcap _k X_k = \varnothing.$$