Application of finite intersection property

Question

If $\mathscr{F}$ be a family of compact sets with finite intersection property in a metric space (X, d). Then show that $\cap \mathscr{F} \neq \phi$.
My thinking: I want to prove by contradiction.
Let, $\cap \mathscr{F} = \phi$ Then {$X-F: F\in \mathscr{F}$} is an open cover of X.
I don't know what to do next. If, I am on the right track, then, please somebody help me. Any independent idea is also welcome. Thanks in advance.

Answer 1

You have the right general idea, but you need to refine it a bit. Pick any $F_0\in\mathscr{F}$, and consider

$$\big\{X\setminus F:F\in\mathscr{F}\setminus\{F_0\}\big\}\;;$$

this is an open cover of the compact set $F_0$. Can you finish it from here?

Answer 2

Brian M. Scott as you have given me the hints, now I'm writing this in response. Please correct me if I am wrong.
Now there will be a finite subcover of $F_0$. Let {$X-F_1$,$X-F_2$,...,$X-F_n$}be that finite subcover. Then $F_0$ $\subseteq$ $\cup (X-F_i) $ for i=1,2,...,n. Which implies $F_0$ $\subseteq$ ($X - \cap F_i$) for i=1,2,...,n. Which implies that $F_0 \cap F_i = \phi$ for i=1,2,...,n. Which contradicts finite intersection property . So, the assumption was wrong.

I'm sorry for I'm writing this in answer. Because after 5 minutes I am not able to put this in comment. I hope this is forgivable.