If I have a compact set $A$ and a closed subset $\Sigma \subset A$ which only contains isolated points (that is, none of them is a limit point). Does the compactness of $A$ then force $\Sigma$ to have finite cardinality ?
Here is my attempt at a proof that the above question can be answered in the positive:
Suppose for contradiction that $\Sigma$ contains infinitely many distinct points.
EDIT : Then we can construct a sequence of points in $\Sigma$ which consists of distinct points.
By compactness of A, this sequence must have a convergent subsequence, and by the fact that $\Sigma$ is closed, this limit lies in $\Sigma$. But then it cannot be a limit point, because all points in $\Sigma$ are isolated. So the subsequence must eventually constant and equal to the limit, contrary to the construction of the sequence.
Is the reasoning above correct ? If no, what did go wrong ?