Showing that intersection of arbitrarily many compact sets in Hausdorff space is compact.

Question

For a topological space $X$ that is Hausdorff, and $\{B_k\}_{k\in I}$ is an arbitrary collection of compact sets, I need to show that $\bigcap\limits_{k\in I} B_k$ is also compact.

Here's the rub. I do not have closed sets to work with. The proof that compact sets in a Hausdorff space are closed is proven in the next section of the book, so I am forbidden from using it in my proof. Unfortunately, I cannot find a proof online anywhere that doesn't rely on closedness. And I've been struggling with this proof for three weeks. Any pointers or full fledged proofs that avoid closedness would be appreciated. I thought I could prove this on my own but have only failed. Im out of time and even when cheating I couldn't find a single useful bit of advice.

Edit A few people have offered some advice, which I appreciate. But to be clear, my instructor isn't going to expect much more beyond the open set cover definition of compactness. We really have little else to work with, sorry to say. I don't know what duals are, I cannot infer that compact sets are closed, and I'm sure there are lots of tricks up people's sleeves that I am just not allowed to use at this juncture.

Answer 1

The first part is taken from Brian M. Scott in How to prove that a compact set in a Hausdorff topological space is closed?.

Lemma: Let $B$ and $C$ be two compact sets in $X$ and fix $x\in B\setminus C$. Then there exists an open set around $x$ that is disjoint from $C$.

Proof: For each $y\in C$ we can, by Hausdorff, construct two disjoint open sets $U_y$ and $V_y$ such that $x\in U_y$ and $y\in V_y$. Now $\left\{V_y\mid y\in C\right\}$ is an open cover of $C$, so by compactness there exists a subcover $\left\{V_y\mid y\in \tilde C\right\}$, where $\tilde C \subset C$ is finite. Now $$ \bigcap_{y\in \tilde C} U_y $$ is an open set that contains $x$ and is disjoint from $C$.

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Let $Y = \bigcap_{k\in I} B_k$ and let $\mathcal{O}$ be an open cover of $Y$. Now, for each $x \in B_1\setminus Y$ there exists a $k\in I$ such that $x\notin B_k$. Therefore, by the lemma, there exists an open set $O_x(k)$ such that $x\in O_x(k)$ and $O_x(k) \cap Y \subseteq O_x(k)\cap B_k = \emptyset$. We then have an open cover of $B_1$ given by $$ \mathcal{O} \cup \left\{ O_x(k)\mid x\in B_1\setminus Y\right\}, $$ which by the compactness of $B_1$ has a finite subcover. Since any $y\in Y$ is not covered by any element in $\left\{ O_x(k)\mid x\in B_1\setminus Y\right\}$ we conclude that there must be a finite subset of $\mathcal{O}$ that covers $Y$.