First-countable implies compactly generated

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.