= Solution
A <filter on a set> $X$ is a nonempty family $\mathcal F\subseteq\mathcal P(X)$ that excludes the empty set, is upward closed, and is closed under finite intersections. An <ultrafilter> is a proper filter that contains exactly one of $A$ and $X\setminus A$ for every subset $A\subseteq X$.
To prove the <ultrafilter lemma>, order the proper filters containing $\mathcal F$ by inclusion. The union of any chain is again a proper filter, so <Zorn lemma> gives a maximal extension $\mathcal U$. If neither $A$ nor $X\setminus A$ belonged to $\mathcal U$, adjoining either one would generate an improper filter. There would then be $U,V\in\mathcal U$ with $U\cap A=\varnothing$ and $V\cap(X\setminus A)=\varnothing$. But $U\cap V=\varnothing$, contradicting propriety. Hence $\mathcal U$ is an ultrafilter.
The <Stone-Čech compactification of the natural numbers> $\beta\mathbb N$ is the set of all ultrafilters on $\mathbb N$, with basic sets
$$
\overline A=\{\mathcal U:A\in\mathcal U\},
\qquad A\subseteq\mathbb N.
$$
The identities
$$
\overline A\cap\overline B=\overline{A\cap B},
\qquad
\beta\mathbb N\setminus\overline A=\overline{\mathbb N\setminus A}
$$
show that these sets form a basis of <clopen sets>. Distinct ultrafilters disagree on some $A$; one lies in $\overline A$ and the other in the disjoint set $\overline{\mathbb N\setminus A}$. Thus $\beta\mathbb N$ is a <Hausdorff space>.
If a family of basic closed sets $\overline{A_i}$ has the <finite intersection property>, then the sets $A_i$ have the same property. They generate a proper filter, which extends to an ultrafilter lying in every $\overline{A_i}$. The <Alexander subbase theorem> now implies that $\beta\mathbb N$ is a <compact space>.
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