As a step in a proof I've been trying to show that convergence of a sequence implies it must have precisely one accumulation point. This is the definition we use for an accumulation point
Let $S$ be a set of real numbers. A real number is an accumulation point $s_0$ of $S$ if and only if for any $\epsilon > 0$, there exists at least one point $t$ of $S$ such that $0 < |t-s_0| < \epsilon$.
I wish to prove:
Lemma: A convergent sequence has precisely one accumulation point.
My thoughts: Since the limit of the sequence exists we know that the set cannot have 2 accumulation points or more, we will show by contradiction.
If we would have multiple accumulation points the limit does not exist because we can never get arbitrarily close to a single point (within ϵ), because there exist certain subsequences that each get arbitrarily close to at least two accumulation points, which determine the minimum distance a sequence can be from an accumulation point (so we get a lower bound and therefore we do not get convergence). There must be precisely one accumulation point.
I do not know how to make this more precise of a statement, there are a lot of words and it's not really structured. I'm looking for some help making a more rigourous argument.