I would like to prove the following statement.
If $A$ is a bounded, infinite subset of $\mathbb{R}$, then there is an element $a \in A$ such that $A-\{a\}$ contains a sequence which converges to $a$.
I found this statement in an introductory course in logic which suggested to use the axiom of choice.
My approach: First I would like to find a non-isolated point in $A$, let's call it $a$. It seems that such a point must exist since only finitely many isolated points fit in the bounded region (I guess). Afterwards, we can choose (with the axiom of choice) a point in every interval around $a$ with the length of the interval going to zero. The sequence of those points has to converge to $a$.
Yes, your application of the axiom of choice is generally right, but the "fine print" is wrong. But this doesn't mean that you got a sequence, because you have $2^{\aleph_0}$ intervals around $a$.
Instead you want to choose from the sets $(a-\frac1n,a+\frac1n)\cap A$, or something like that. Because a sequence is a function from $\Bbb N$ into $A$. So you need to be able to tell which element is the first, second, and so on. You also want them to be from $A$, rather than just from an interval around $a$.
This would also be a good time to indicate that the axiom of choice is necessary here, and it is consistent without the axiom of choice that there is a set of real numbers $D$ which has no isolated points, but every convergent sequence of elements of $D$ is eventually constant. In that case picking any $d\in D$ we have that $D\setminus\{d\}$ contains no sequence converging to $d$.