Sometimes it is difficult to know whether or not we are formally invoking the axiom of choice. I wonder if some basic math proofs, which are not ordinarily associated with the axiom of choice, are in fact “secretly” using the axiom of choice?
Consider for example the following standard fact of linear algebra:
Suppose $V$ and $W$ are vector spaces, each with an uncountably infinite number of vectors. Let $T:V\rightarrow W$ be a surjective linear map.
Claim: If $\{v_1, …, v_n\}$ spans $V$, then $\{T(v_1), …, T(v_n)\}$ spans $W$.
Proof: Fix $w \in W$. Since $T$ is surjective, there exists a $v_w \in V$ such that $T(v_w)=w$, [and so on with the standard proof]. $\Box$
The idea here is that the proof must hold for all $w \in W$. So we must choose a $v_w \in V$ for each of the uncountably many $w \in W$. Is this an implicit use of the choice axiom?
No, that argument does not depend on AC. You can argue that it "morally" needs infinitely many choices, but it does that while proving a claim, not while constructing an object. Or, in other words, nothing that depends on the choices it makes needs to be remembered beyond "yes, this is true for $w$ too".
I've written a longer answer previously that tries to explain some of the technical details that lead to this.
By the way, I think it is somewhat misleading to speak about AC "leaking into" everyday mathematics. The overwhelming consensus among mathematicians these days is that arguments that depend on AC are perfectly fine in "everyday mathematics". Caring about whether a particular argument depends on AC is essentially only done in the specialized area of set theory.