Let $K_1, K_2,\ldots$ be compact sets in a topological space such that $\cap_i K_i = \emptyset$. I want to show that $\cap_i^m K_i = \emptyset$ for some $m$.
But I am struggling to see why because the topological space itself say we call it $X$ need not be compact itself hence we cannot obtain the finite cover of open sets we need. Because we easily see that if we assume the given we get $$\left(\cap_i K_i \right)^c = X \iff \cup_i K_i^c = X,$$ hence we have found an open cover of $X$ but we cannot take a finite amount of $K_i$ such that it still holds as $X$ may not be compact. Any help is greatly appreciated.
Hint. Let $L_i = \bigcap_{j=1}^i K_j$, then the $K_1 \setminus L_i$ form an open cover of $K_1$. Now use $K_1$'s compactness to find an $i$ such that $K_1 = K_1 \setminus L_i$ (note that the $K_1 \setminus L_i$ are an increasing sequence), hence $L_i = \emptyset$.