Maurice Frechet's 1904 Definitions of Compactness 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.


 A: 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 $T$.
