Let $X$ be a topological space. Prove $X$ is quasi-compact iff for any family $(V_i)_{i\in I}$ of closed subset of $X$ with $\bigcap_{j\in J}V_j\neq \emptyset$ for any $J\subseteq I$ with $J$ finite it follows that $\bigcap_{i\in I}V_i \neq \emptyset$
This confuses me, since I have closed sets here. Shouldn't $\bigcap_{i\in I}V_i =\emptyset$?
Because if $X$ is quasi-compact, there exist for every open sets $X=\bigcup\limits_{i\in I}U_i$ a finite $J\subset I$ such that $X=\bigcup\limits_{i\in J}U_i$. This means $$\emptyset =(\bigcup\limits_{i\in J}U_i)^c=\bigcap\limits_{i\in J}U_i^c=\bigcap\limits_{i\in J}V_i$$ And because $\bigcap\limits_{i\in I}V_i=\bigcap\limits_{i\in J}V_i \cap \bigcap\limits_{i\in I}V_i$ we have that $\bigcap\limits_{i\in I}V_i=\emptyset$
This confuses me and I don't know how to solve the problem.
If you assume that $\bigcap_{i\in I}V_i=\varnothing$, you get a contradiction. Specifically, for each $i\in I$ let $U_i=X\setminus V_i$. Then each $U_i$ is open, and
$$\bigcup_{i\in I}U_i=\bigcup_{i\in I}(X\setminus V_i)=X\setminus\bigcap_{i\in I}V_i=X\setminus\varnothing=D\;,$$
so $\{U_i:i\in I\}$ is an open cover of $X$. Therefore there is a finite $J\subseteq I$ such that $\bigcup_{i\in J}U_i=X$; can you finish it by showing that this implies that $\bigcap_{i\in J}V_i=\varnothing$ and thereby getting a contradiction?