Family of Closed set of $X$ having finite intersection property with arbitary intersection is non empty Then $X$ is compact metric space

Question

I wanted to prove Family of Closed set of $X$ having finite intersection property with arbitary intersection is non empty Then $X$ is compact metric space
I wanted to check my argument
On Contrary suppose $(X,d) $ is not compact space ,SO there exist open cover of X but which does not have finite subcover which can cover that X
SO $X=\cup_{a=1}^\infty G_a$ So it does not have finite subcover of this family which can cover X
That is $\exists x\in X $ such that $x\notin \cup_{a=1}^n G_a $ for some n....(1)
As $G_a$ is open $\to$ X\ $G_a=G_a'$ is closed.[Notation for complement ']
So here $\cap^n_1 G'_a\neq \phi$ as $x\in\cap^n_1 G'_a $ Form 1
Which implies $\cap^\infty_1 G'_a\neq \phi$ by assumption which means
$x\in \cap^\infty_1 G'_a$ i.e $x\notin X$as covering assumption .
Now this is contradition to assumption that every $x\in X$as by construction of $G_a$
How was my argument ? Please tell me .

Answer 1

Your argument is muddled.

Clarify it:

We assume that $X$ has the property $\ast$ that a family of closed sets with the finite intersection property has non-empty intersection. Suppose (for a contradiction) that $X$ is not compact. Then there is an open cover $U_i, i \in I$ of $X$ that has no finite subcover.

Define $F_i = X\setminus U_i$. Then all $F_i$ are closed. Show that every finite intersection of $F_i$ has non-empty intersection. Hint: use that

$$\cap_{i \in F} F_i = \cap_{i \in F} (X\setminus U_i) = X \setminus (\cup_{i \in F} U_i)$$

for any (finite) subset $F$ of $I$.

By the fact that $X$ has property $\ast$, we know that $\cap_i F_i \neq \emptyset$. Show that an element of this intersection would not be covered by any $U_i$. This contradiction then shows that the assumption that $X$ is not compact was wrong.

So $X$ is compact.