Here is Prob. 5, Sec. 28, in the book Topology by James R. Munkres, 2nd edition:
Show that $X$ is countably compact if and only if every nested sequence $C_1 \supset C_2 \supset \cdots$ of closed nonempty sets of $X$ has a nonempty intersection.
Here is a proof of this result.
My Attempt:
Let $X$ be a topological space.
Suppose that $X$ is countably compact. Let $\left( C_n \right)_{n \in \mathbb{N} }$ be a sequence of nonempty closed sets of $X$ such that $$ C_1 \supset C_2 \supset \cdots. \tag{1} $$ We need to show that $$ \bigcap_{n = 1}^\infty C_n \neq \emptyset. \tag{2} $$
Let us suppose that (2) does not hold. That is, let us suppose that $$ \bigcap_{n = 1}^\infty C_n = \emptyset. \tag{3} $$
For each $n \in \mathbb{N}$, let us put $$ U_n \colon= C_n^\mathrm{c} = X \setminus C_n. \tag{Definition 1} $$ Then each set $U_n$ is open in $X$ and is a proper subset of $X$ since each set $C_n$ is a closed, nonempty subset of $X$. Now by one of the DeMorgan's laws and also using our supposition of (3), we find that $$ \bigcup_{n=1}^\infty U_n = \bigcup_{n=1}^\infty \left( X \setminus C_n \right) = X \setminus \bigcap_{n=1}^\infty C_n = X \setminus \emptyset = X. $$ Thus the collection $$ \left\{ \ U_n \ \colon \ n \in \mathbb{N} \ \right\} $$ is a countable open covering of $X$, and since by our supposition $X$ is countably compact, some finite subcollection of this collection also covers $X$; that is there exists a natural number $N$ such that $$ X = \bigcup_{n=1}^N U_n.$$ But this together with (Definition 1) implies that $$ \bigcap_{n = 1}^N C_n = \bigcap_{n=1}^N \left( X \setminus U_n \right) = X \setminus \bigcup_{n=1}^N U_n = X \setminus X = \emptyset. $$ But (1) implies $$ \bigcap_{n = 1}^N C_n = C_N.$$ Therefore we obtain $$ C_N = \emptyset, $$ which is a contradiction to our choice of the sets $C_n$ as being nonempty closed subsets of $X$. Thus (3) cannot hold. So (2) must hold.
Hence if $X$ is countably compact, then for every sequence $\left( C_n \right)_{n \in \mathbb{N} }$ of nonempty closed sets in $X$ such that $$ C_1 \supset C_2 \supset \cdots, $$ we must have $$ \bigcap_{n=1}^\infty C_n \neq \emptyset, $$ as required.
Conversely, let us suppose that every nested sequence of nonempty closed sets in $X$ has a nonempty intersection. We show that then $X$ is countably compact.
Let $\left\{ \ U_n \ \colon \ n \in \mathbb{N} \ \right\} $ be a countable open covering of $X$.
For each $n \in \mathbb{N}$, let us put $$ V_n \colon= \bigcup_{j = 1}^n U_j. \tag{Definition 2} $$ Then each set $V_n$ is an open set in $X$. Moreover, since $$ \bigcup_{n =1 }^\infty U_n = X, $$ and since $$ U_n \subset V_n \subset X $$ for each $n \in \mathbb{N}$, therefore we note that $$ \bigcup_{n = 1}^\infty V_n = X \tag{4} $$ also. Thus the collection $\left\{ \, V_n \colon n\in \mathbb{N} \, \right\}$ is also an open covering of $X$.
If one of the sets $V_n$, say the set $V_N$, equals $X$, then the finite subcollection $\left\{\ U_1, \ldots, U_N \ \right\}$ of the open covering $\left\{ \ U_n \ \colon \ n \in \mathbb{N} \ \right\} $ covers $X$.
So let us assume that the set $X$ itself is not in the collection $\left\{ \ V_n \ \colon \ n \in \mathbb{N} \ \right\} $ of open sets of $X$.
For each $n \in \mathbb{N}$, let us put $$ C_n \colon= V_n^\mathrm{c} = X \setminus V_n. \tag{Definition 3} $$ As each set $V_n$ is open in $X$, so each set $C_n$ is closed.
Now by our assumption and also by virtue of (Definition 2) above, we note that, for each $n \in \mathbb{N}$, we have $$ V_n \subset V_{n+1} \subsetneqq X,$$ from which (upon taking complements) we obtain $$ X \setminus V_n \supset X \setminus V_{n+1} \supsetneqq X \setminus X,$$ which by (Definition 3) above is the same as $$ C_n \supset C_{n+1} \supsetneqq \emptyset. $$ Thus $\left( C_n \right)_{n \in \mathbb{N} }$ is a nested sequence of nonempty closed sets of $X$. Therefor by our hypothseis we have $$ \bigcap_{n=1}^\infty C_n \neq \emptyset, $$ and so $$ \bigcup_{n = 1}^\infty V_n = \bigcup_{n = 1}^\infty \left( X \setminus C_n \right) = X \setminus \bigcap_{n = 1}^\infty C_n \subsetneqq X.$$ But this contradicts the fact that the collection $\left\{ \ V_n \ \colon \ n \in \mathbb{N} \ \right\}$ covers $X$. So our assumption that no $V_n$ equals $X$ is impossible. Hence some $V_N$ equals $X$, and thus $X$ is covered by $\left\{ \ U_1, \ldots, U_N \ \right\}$. Refer to (Definition 2) above.
Since $\left\{ \ U_n \ \colon \ n \in \mathbb{N} \ \right\}$ was an arbitrary countable open covering of $X$, we can conclude that every countable open covering of $X$ has a finite subcollection also covering $X$. Hence $X$ is countably compact.
Is my presentation of the proof any clearer than the proof given here?
Your proof is correct, all details are there.
In real life such arguments are of course presented much more succinctly (one just says : "obvious by taking complements"), as the translation between open covers and intersections of closed sets is considered pretty standard in topology contexts, as is the trick of having increasing covers and decreasing closed sets, both via complementation (and de Morgan's laws). So whether one is clearer or not depends on the context. In a learning environment it can be good to spell out everything to ensure no details are swept under the rug, as it were. In a paper this would be overkill, and left to the reader.