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. 

We do not know that $a_n\leq a^*$. For example, $\frac1n\leq 0$ is not true, even though $0$ is the limit of $\frac1n$.
Remember, $\sup A$ is defined as the smallest upper bound for $A$. This means you need to prove that 


*

*$\sup A$ exists (i.e., $A$ is bounded)

*For every $\epsilon > 0$, $A^*-\epsilon$ is not the supremum.

