Elementary theorems that require AC It seems that AC is hiding (maybe concealed?) even in some elementary results.
An example:

Theorem: Let $X \subseteq \mathbb R$ and let $x_0 \in \mathbb R$ be an accumulation point of $X$. Then  there exists a sequence $ \{ a_n \}_{n=1}^\infty $ S.T.  $ \{ a_n \} \subseteq X$ and $a_n\xrightarrow{n \to \infty} x_0 $.
Proof: For $\mathbb N \ni n > 0$ we denote $A_n := \{ x \in X :  |x-x_0| < \frac {1}{n} \}$, since $x_0$ is an accumulation point of $X$ then $ \forall [ 0<n \in \mathbb N ] . A_n \neq \varnothing$.

By A.C. there exists choice function $f:P(X) \setminus \{ \varnothing\} \rightarrow X$ S.T. $\forall [ \varnothing\neq B \subset X] . f(B) \in B$

The sequence $\{ a_n \}$ defined by $a_n := f(A_n)$ satisfies the requirements.


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*Can we avoid the use of AC in the Theorem above??

*Can you point out some elementary Theorems that require AC?


As pointed out the theorem seem to require the AC, but there seem to be a way around this (in this particular case) by changing definition to:

$x$ is a accumulation point to $X$ if there is a sequence $x_n\in X$ such that $lim_{n\to\infty}x_n = x$.

Then of course the theorem will be trivial to prove. The reverse (that every neighborhood of $x$ intersects $X$) is provable (since the sequence $x_n$ must eventually be confined to any given neighborhood of $x$).
I guess that this is the case for all "elementary" theorem that requires the AC that they can be "solved" by shifting a definition to a version that is equivalent under AC, but only implying the original defining property without AC. In most "elementary" applications to the theory you would probably have an explicit way of doing your choice that's required for the modified definition so you can get away without AC.
