One may prove that the Axiom of Choice is equivalent to the following statement $P$:
If $\{(X_i,\tau_i)\mid i\in I\}$ is a system of topological sets, and $\prod_{i\in I}X_i$ is equipped with the corresponding product topology, then $$\overline{\prod_{i\in I}A_i}=\prod_{i\in I}\overline{A_i}$$ whenever $A_i\subseteq X_i$ for each $i\in I$.
I am looking for an interesting result whose proof relies on a direct application of $P$ (and which wouldn't be provable without assumption of $P$, or AC).
To give a different example, the Axiom of Choice is also equivalent to Tychonoff's Theorem. An application of Tychonoff's Theorem results in the De Bruijn–Erdős Theorem.
If you assume that this theorem holds (in some model of set theory), it follows that AC holds too in that model. The proof is probably based on defining a product of topological spaces and subsets $A_i$ for a family of set that needs a choice function. I don't recall ever seeing this proof myself, it's probably not too hard; I do show in this note how the Tychonoff theorem (a product of compact topological spaces is compact) also proves AC; I'd imagine the proof of "closures result implies AC" to be something like it.
So then you can use AC to prove all other results that depend on AC, like the De Bruijn–Erdős theorem that you mentioned. Any application of AC could be seen as an application of this result.
As AC can be proved from the closures result, it automatically follows that it cannot be shown without AC, as AC is independent from the other ZF axioms.