First-countable implies compactly generated

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., first-countable space) must also satisfy the second topological space property (i.e., compactly generated space)
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Statement

Property-theoretic statement

The property of topological spaces of being first-countable is stronger than the property of being compactly generated.

Verbal statement

Any first-countable space is compactly generated.

Proof

Given: A first-countable space $X$

To prove: There exists a collection of compact subsets $\{K_{i}\}_{i\in I}$ of $X$, such that $U\subset X$ is open if and only if $U\cap K_{i}$ is open in $K_{i}$ for every $i$

Construction of the collection of compact subsets

We consider compact subsets of the following form: take a sequence $x_{n}$ that converges to a point $x$ (note that a sequence may converge to more than one point, we just make sure there is at least one point of convergence). Now define the compact set $K$ corresponding to this sequence as $\{x_{n}\}_{n\in \mathbb {N} }\cup \{x\}$.

Let's see why $K$ is compact. First, observe that by construction, any open set of $K$ containing $x$ must contain all but finitely many $x_{n}$. Hence, if we have an open cover of $K$, the member of the open cover containing $x$ must contain all but finitely many of the $x_{n}$s. Picking one member of the open cover for each $x_{n}$, we obtain a finite subcover.

Proof that this collection works

Given: A subset $V$ of $X$ such that $V\cap K$ is open in $K$ for every $K$ of the above form.

To prove: $V$ is open in $X$

Proof: We want to show that for every $x\in V$, there exists an open subset of $X$ containing $x$ and inside $V$.

Since $X$ is first-countable, we can find, for any point $x\in X$, a descending chain $U_{n}$ of open subsets containing $x$, such that every open set containing $x$ contains one of the $U_{n}$s. It thus suffices to show that one of the $U_{n}$s is contained in $V$.

Suppose not. Then, none of the $U_{n}$s is contained in $V$. Pick $a_{n}\in U_{n}\setminus V$ for each $n$, and consider the sequence of $a_{n}$s. Any open subset containing $x$ contains one of the $U_{n}$s, so it contains all but finitely many of the $a_{n}$s. Thus, the $a_{n}$s converge to $x$ Consider the compact set $K$ formed by the $a_{n}$s and $x$. By assumption, $V\cap K$ is open in $K$. But this would force $V\cap K$ to contain one or more of the $a_{n}$s, a contradiction. This completes the proof.