I'm writing a small paper on the history of compactness.
Frechet wrote in French, and I don't speak French, so I've been consulting this paper: Taylor, A.E.
On page 244, I read that Frechet proved that the intersection of any nested sequence of closed subsets of a limit-point compact space is non-empty, and conversely.
Why is this true?
I believe Frechet's spaces are called Frechet-Urysohn spaces today (cf. Sequential Spaces).
Here is an excerpt from Frechet's 1906 thesis in which he discusses the topic.
For the one direction, we can use the original proof, as contained in the excerpt. Using the notation of the original, but not clinging to the formulation (which I could only very clumsily attempt to translate),
Let $E$ be a compact (limit point compact) set, and $E_1 \supset E_2 \supset \dotsc \supset E_n \supset \dotsc$ a nested sequence of nonempty closed subsets of $E$. There is at least one element $M_1 \in E_1$, $M_2 \in E_2$ etc.
If there are infinitely many distinct elements among the $M_n$, let these be $M_{n_1},\, M_{n_2},\,\dotsc$. Since $E$ is (limit point) compact, this last sequence has at least one limit point $M \in E$. Since the sequence is contained in $E_{n_p}$ from the term $M_{n_p}$ on, $M$ is a limit point of $E_{n_p}$, and since $E_{n_p}$ is closed, $M \in E_{n_p}$, and hence in $E_1,\, E_2,\, \dotsc,\, E_{n_p}$. Since that holds for all $p$, $M$ is a common point of all $E_n$, i.e. $M \in \bigcap\limits_{n=1}^\infty E_n$.
If on the other hand, the sequence $M_1,\, M_2,\, \dotsc$ contains only finitely many distinct terms, there is at least one element $M$ that is repeated infinitely often in the sequence, $M = M_{n_1} = M_{n_2} = \dotsc$. Therefore, $M \in E_{n_p}$, and hence in $E_1,\, E_2,\, \dotsc,\, E_{n_p}$, for all $p$, therefore $M$ is a common element of all $E_n$, i.e. $M \in \bigcap\limits_{n=1}^\infty E_n$.
For the converse, I unfortunately cannot draw on Fréchet's proof directly, so I'll have to cook up my own version of the standard proof.
Let $E$ be a topological space such that every nested sequence of closed subsets has nonempty intersection. We want to show that every infinite subset of $E$ has at least one limit point in $E$.
Let $T \subset E$ be an infinite subset. Let $S \subset T$ be a countably infinite subset, and $(s_n)_{n\in \mathbb{N}}$ an enumeration of $S$.
For $n \in \mathbb{N}$, let $E_n = \overline{\{s_k : k\geqslant n\}}$. Then $E_n$ is a nonempty closed subset of $E$, and $E_n \supset E_{n+1}$ for all $n$, hence $(E_n)$ is a nested sequence of nonempty closed subsets. By assumption, there is an $M \in \bigcap\limits_{n\in\mathbb{N}} E_n$. This is a limit point of $S$, and hence of $T$: let $U$ be any neighbourhood of $M$. Since $M \in E_1$, there is an $n_1 \geqslant 1$ with $s_{n_1} \in U$. Since $M \in E_{n_1+1}$, there is an $n_2 > n_1$ with $s_{n_2} \in U$. If the definition of limit point is that every neighbourhood contains at least one point from the set distinct from $M$, we're done now, since $s_{n_1} \neq s_{n_2}$, and at least one of the two is therefore distinct from $M$. If the definition of limit point is that every neighbourhood contains infinitely many points of the set (which nowadays would be called an accumulation point; for $T_1$ spaces, both notions coincide), we continue in the above way to construct a subsequence $(s_{n_k})$ of $(s_n)$ with $n_{k+1} > n_k$ and $s_{n_k} \in U$ for all $k$. Thus $U$ contains infinitely many points of $S$, hence of $T$, and since $U$ was arbitrary, $M$ is a limit point (accumulation point) of $S$, and hence of $T$.