Partition of unity without AC

Several existence theorems for partition of unity are known. For example (source),

Proposition 3.1. If $$(X,\tau)$$ is a paracompact topological space, then for every open cover $$\{U_i \subset X\}_{i \in I}$$ there is a subordinate partition of unity (def. 2.4).

The axiom of choice is used in the proof of this theorem.

Is there an existence theorem for partition of unity that can be proved without the axiom of choice (including variants such as countable choice, dependent choice, etc)?

For example, what if we strengthen the assumption to separable locally compact metric space?

• For separable metric spaces, there's probably a Shoenfield absoluteness argument to remove choice from the proof. More generally, something similar probably works for any metric space with a dense well-ordered subset. (That said, if this does work, a direct proof is probably not that bad in the first place.) Commented May 17 at 7:16

• Is there no chance to avoid Urysohn's lemma? For metric spaces, the existence of a function separating two given disjoint closed sets can be proved without AC. ( $\frac{d(x, A)}{d(x, A)+d(x, B)}$ is $0$ on $A$ , $1$ on $B$.) Commented May 17 at 10:31