Infinite Intersection of Descending Sequence of Non-empty, Compact Subsets. I'm working on a problem that asks me to prove that the infinite intersection of a nested sequence of non-empty compact sets is non-empty. I thought I proved it, and my professor more-or-less agreed, except that he says I missed a crucial element. I proved it as follows:
Let $(x_n)$ be a sequence such that $x_1 \in K_1$, $x_2 \in K_2$, $\cdots$, $x_i \in K_i$. Such a sequence exists because $K_n \ne \{\emptyset\} \ \forall n$.
Thus, because $K_1$ is compact, and $(x_n) \subset K_1$, $(x_n)$ has some convergent subsequence such that $(x_n) \to x$. Moreover, by construction of $(x_n)$, $x \in K_n \ \forall n$. Thus, $x \in \lim_{n \to \infty} K_n$. We thus conclude:
$$x \in \bigcap_{n=1}^\infty K_n $$
And thus, $\bigcap K_n \ne \{\emptyset\}$
My professor agreed with this proof, but added that it was important to note that $F_i$ is closed for any $i$, because $K_1$ is compact. Clearly, this is a true statement, but I don't understand why this is a crucial part of the proof. Was me misunderstanding me, or is there something lacking in my proof as is?
 A: I think we can state the following:

Let $X$ be any Hausdorff topological space, and let $K_1, K_2, K_3, \ldots$ be a collection of non-empty sequentially compact subspaces of $X$ such that
$$
K_1 \supset K_2 \supset K_3 \supset \cdots. 
$$
Then the intersection $K_1 \cap K_2 \cap K_3 \cap \cdots$ is non-empty.

Proof:

For each $n = 1, 2, 3, \ldots$, let $x_n \in K_n$. This assumption is reasonable because each $K_n$ is non-empty.


Since $\left( x_n \right)_{n \in \mathbb{N}}$ is a sequence in $K_1$, there is a subsequence $\left( x_{1, n} \right)_{n \in \mathbb{N}}$ of $\left( x_n \right)_{n \in \mathbb{N}}$ converging to some point $x \in K_1$.


Since (some tail of) the sequence  $\left( x_{1, n} \right)_{n \in \mathbb{N} }$ is a sequence in the sequentially compact space $K_2$, some subsequence $\left( x_{2, n} \right)_{n \in \mathbb{N}}$ converges in $K_2$, and since $\left( x_{1, n} \right)_{n \in \mathbb{N} }$ converges to $x$, we can conclude that $x \in K_2$ also.


We can continue this pattern of reasoning to conclude that our point $x$ is in every $K_m$ for $m = 1, 2, 3, \ldots$.

