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A choice function $f$ is defined on a collection $X$ of nonempty sets and $\forall A \in X $ we have $f(A)\in A$. So the Axiom of choice states that for any set $X$ of nonempty sets, there exists a choice function $f$ defined on $X$.

Could you please explain (with examples, if possible) why the Axiom of choice leads to non-constructive proofs?

Thank you for your help

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    It's literally in the definition. "There exists a choice function". When using the AOC in proofs, we can say there exists a choice function (or equivalently a well-order) without explicitly specifying it. – 2017-02-18
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    I think that I minced many words on this topic, right here on this website. Have you looked around? – 2017-02-19
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    To expand on MathematicsStudent1122, if you can explicitly construct the choice function you need, then you didn't need the axiom of choice to tell you it exists. If you need the axiom of choice, then you don't have a way to construct $f$. You just know that some such $f$ must exist. But you don't know what it is. Any other object whose existance you deduce from $f$ similarly is unknown, because your construction of it depends on the unknown values of $f$. – 2017-02-19

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