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.


*

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

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


Yes, this proof uses countable choice. In an essential way, too. It is consistent (without choice) that there is a dense set if reals without a countably infinite subset. In particular every convergent sequence from that set must be eventually constant. But density means that every real is in the closure.
Other proofs that use the axiom of choice include:


*

*Every infinite set has a countably infinite subset; 

*a set is finite if and only if every self-injection is a bijection; 

*the countable union of countable sets is countable; and

*an infinite tree without maximal nodes, where every level is finite, has a branch.


Slightly less elementary proofs might include 


*Every vector space has a basis; and 

*a vector space is infinite dimensional if and only if its [algebraic] dual has a larger dimension.


The list is really quite large and can fill up several books.
