Showing that an arbitrary intersection of compact sets is compact in $\mathbb{R}$

Question

Claim: For any collection of compact sets $K_n$ where $n \in \mathbb{N}$ the arbitrary intersection $\cap_{n \geq 1}K_{n}$ is also compact.

Attempt: To show compactness I am going to appeal to the Heine-Borel Theorem and show closed and boundedness.

Closed: I've shown previously that a finite or infinite intersection of closed sets is closed so this would suffice for this portion.

Bounded: This is where I am having trouble showing it. It intuitively makes sense to me that an intersection of bounded sets will also be bounded, but trying to write this out formally is giving a bit of trouble. I thought perhaps contradiction:

Assume that the intersection is not bounded, but given that each set individually is bounded then there are individual bounds on those elements and an intersection of each of those sets of elements will still be bounded by their previous respective bound. So the intersection must also be bounded.

Would this be the correct reasoning or am I missing something?

Answer 1

If a set $A$ is bounded, then $A$ is contained in some ball $B(x,r)$. Therefore, every subset of $A$ is contained in $B(x,r)$. Therefore, the intersection of $A$ with anything whatsoever is contained in $B(x,r)$ and so is bounded, because the intersection of $A$ with anything whatsoever, whether it be 1 other set or a trillion other sets or infinitely many other sets, is a subset of $A$.

Answer 2

Given compact subsets $K_i$, $i\in I$, of a Hausdorff topological space $X$. Assume that $I\ne \emptyset$, say, $i_0\in I$. Let $A=\bigcap_{i\in I} K_i$. Let $\{U_j\}_{j\in J}$ be an open cover of $A$. We know that the $K_i$ are closed. Then $\{U_j\}_{j\in J}\cup \{K_i^\complement\}_{i\in I}$ is an open cover of $K_{i_0}$ (in fact, it covers all of $X$). A finite subcover of this consists of finitely many $U_j$ and finitely many $K_i^\complement$. As the latter are disjoint from $A$, we have found finitely many $U_j$ that cover $A$.