Is this proof correct? i'm new at this, so i´m really sorry sorry if i do something wrong.
The problem is this:
A choice function satisﬁes condition α if whenever $x = C(A)$ and $x ∈ B ⊂ A$, it follows that $x = C(B)$ as well.
Axiom 1: If $x ≠ C(A)$ then C(A) = C(A\{x})
And i need to show if condition α implies Axiom 1.

C(A) is choice function. Here is a definition of choice function (just in case you need it): "Let X be a ﬁnite set of alternatives. Let P(X) be the set of all nonempty subsets of X. We call the elements of P(X) as menus. A choice function is a map C : P(X) → X such that C(A) ∈ A, for all A ∈ P(X). It selects a single element from each menu. A choice correspondence is a map C : P(X) →P(X) such that C(A) ⊆ A, for all A ∈P(X). From each menu, at least one alternative is chosen"

And i found that Axiom 1 does not implies Condition α. Here is a counter example: We have X = {x, y, z}. Take A = {x, y, z} and y∈C({x, y, z})
y∈C({x, y}), z∈C({x, z}) and {y, z}∈C({ y, z}). Here we have a contradiction because C({x, y, z})≠C({ y, z}).
Thank you.
Your "counterexample" also violates condition $\alpha$: $y=C(\{x,y,z\})$ but $y\ne C(\{y,z\})$. So it's not really a counterexample.