On a decreasing sequence of compact subsets of a Hausdorff topological space I have the following exercise to solve:

Exercise. Let $ (E,\tau) $ be a Hausdorff topological space and
    $ (K_{n})_{n \in \mathbb{N}} $ a decreasing sequence of non-empty
    $ \tau $-compact subsets of $ E $. Prove that
    $ \displaystyle K \stackrel{\text{df}}{=} \bigcap_{n \in \mathbb{N}} K_{n} $ is
    $ \tau $-compact.
Suppose that there exists a $ \tau $-open subset $ U $ of $ E $ such that
    $ K \subset U $. Then prove that there exists an $ n_{0} \in \mathbb{N} $ such
    that $ K_{n} \subseteq U $ for all $ n \in \mathbb{N}_{\geq n_{0}} $.


To prove that $ K $ is $ \tau $-compact, I mentioned that as the $ K_{n} $’s are $ \tau $-compact, they are $ \tau $-closed, so $ K $ is a $ \tau $-closed subset of $ E $ contained within a $ \tau $-compact subset of $ E $. Therefore, $ K $ is $ \tau $-compact.

However, I don’t know how to solve the second question or how to see that $ K \neq \varnothing $. Could someone please suggest another method to prove that $ K $ is $ \tau $-compact?
Thank you!
 A: This post was re-written because the OP apparently had trouble understanding the original version. The argument should be crystal clear now.

Proof that $ K $ is a compact subset of $ E $.
$ K $ is an intersection of closed subsets of $ E $ (in a Hausdorff space, compact subsets are closed), so $ K $ itself is a closed subset by the axioms of a topology. Then as $ K \subseteq K_{1} $ and $ K_{1} $ is a compact subset, it follows that $ K $ is a compact subset as well (a closed subset of a compact subset is compact). $ \quad \clubsuit $

Lemma. $ K $ is non-empty.
Proof
Assume for the sake of contradiction that $ \displaystyle K = \bigcap_{n = 1}^{\infty} K_{n} = \varnothing $. Then by de Morgan’s Laws, we have
\begin{align}
    K_{1}
& = K_{1} \setminus \varnothing \\
& = K_{1} \bigg\backslash \bigcap_{n = 1}^{\infty} K_{n} \\
& = \bigcup_{n = 1}^{\infty} (K_{1} \setminus K_{n}).
\end{align}
Let $ n \in \mathbb{N} $. Observe that $ K_{1} \setminus K_{n} = K_{1} \cap (E \setminus K_{n}) $, which shows that $ K_{1} \setminus K_{n} $ is the intersection of $ K_{1} $ with an open subset of $ E $. By definition, $ K_{1} \setminus K_{n} $ is thus a $ K_{1} $-open subset of $ K_{1} $.
Hence, $ \mathcal{O} \stackrel{\text{df}}{=} \{ K_{1} \setminus K_{n} \}_{n = 1}^{\infty} $ is a $ K_{1} $-open cover of $ K_{1} $, and as $ K_{1} $ is a compact topological space, we can find natural numbers $ n_{1},\ldots,n_{k} $ such that $ \mathcal{O}' \stackrel{\text{df}}{=} \{ K_{1} \setminus K_{n_{i}} \}_{i = 1}^{k} $ is a finite $ K_{1} $-open subcover of $ K_{1} $. Without any loss of generality, we may assume that $ n_{1} < \ldots < n_{k} $, which yields
\begin{align}
             K_{1}
& =          \bigcup_{i = 1}^{k} (K_{1} \setminus K_{n_{i}}) \\
& =          K_{1} \bigg\backslash \bigcap_{i = 1}^{k} K_{n_{i}} \qquad
             (\text{By de Morgan’s Laws again.}) \\
& =          K_{1} \setminus K_{n_{k}} \qquad
             (\text{As $ K_{n_{k}} \subseteq \ldots \subseteq K_{n_{1}} $.}) \\
& \subsetneq K_{1}. \qquad
(\text{As $ K_{n_{k}} \subseteq K_{1} $ and $ K_{n_{k}} \neq \varnothing $ by hypothesis.})
\end{align}
We thus have the contradiction $ K_{1} \subsetneq K_{1} $, and so our assumption that $ K $ is empty is truly rubbish. Therefore, $ K $ is non-empty. $ \quad \spadesuit $

Theorem. If $ K \subseteq U $ for some open subset $ U $ of $ E $, then there exists an $ n_{0} \in \mathbb{N} $ such that $ K_{n} \subseteq U $ for all $ n \in \mathbb{N}_{\geq n_{0}} $.
Proof
Suppose that $ K \subseteq U $ for some open subset $ U $ of $ E $. Then
\begin{align}
    \varnothing
& = K \setminus U \\
& = \left( \bigcap_{n = 1}^{\infty} K_{n} \right) \bigg\backslash U \\
& = \bigcap_{n = 1}^{\infty} (K_{n} \setminus U).
\end{align}
Let $ n \in \mathbb{N} $. Observe that $ K_{n} \setminus U = K_{n} \cap (E \setminus U) $, which implies that $ K_{n} $ is an intersection of two closed subsets of $ E $. Hence, $ K_{n} \setminus U $ is a closed subset, and as it is at the same time contained inside the compact subset $ K_{n} $, it is a compact subset as well.
We now see that $ (K_{n} \setminus U)_{n = 1}^{\infty} $ is a decreasing sequence of compact subsets of $ E $ whose intersection is empty. The lemma above tells us that if all members of this sequence were non-empty, then their intersection would be non-empty as well. However, this is not the case! Hence, some member of the sequence must be empty, i.e., $ K_{n_{0}} \setminus U = \varnothing $ for some $ n_{0} \in \mathbb{N} $, or equivalently, $ K_{n_{0}} \subseteq U $. Then as
$$
          \ldots
\subseteq K_{n_{0} + 3}
\subseteq K_{n_{0} + 2}
\subseteq K_{n_{0} + 1}
\subseteq K_{n_{0}}
\subseteq U,
$$
we conclude that $ K_{n} \subseteq U $ for all $ n \in \mathbb{N}_{\geq n_{0}} $. $ \quad \blacksquare $
