Supremum proof simple I got stuck on this problem and can't figure it out, I hope somebody can help me, I also wrote my attempt. Thanks in advance!!

Question: Let $(a_n)$ be a convergent sequence in $\mathbb{R}$. $a_n
 \to a^*$. Let $A=\{a_n | n \in \mathbb{N}\}$. I have to show that:
  $\sup A  \geq a^*$

My attempt:
Suppose $a_n \in A$.
$a_n$ is bounded because it is convergent. Because $A=\{a_n | n \in \mathbb{N}\}$ we can say that $\forall a_n \in A$ : $a_n \leq \sup A$.
We also know that $a_n \leq a^*$, because it's the limit. 

It is easy to show that the supremum exists so I'll leave that for you to show (you've pretty much done it anyway).
Now suppose by way of contradiction that $supA<a^{*}$.
Then $\exists  \epsilon_{0}>0$ such that $|a_{n}-a^{*}|\geq\epsilon_{0}\, \forall n \in N$ (follows almost directly from the definition of the supremum). This contradicts the fact that the sequence converges to $a^{*}$ (you should show this using the definition of convergence).
