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. 

Let $b$ be an upper bound for $A$.
Then $a_n \le b$ for all $n$, and so $\displaystyle a^*=\lim_{n\to\infty} a_n\le b$.
Since $\sup A$ is an upper bound for $A$, we can take $b=\sup A$ and conclude that $a^* \le \sup A$.
