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Epsilon proof of an open set.
This is an easy question, I know it but I’m not sure how to approach it.
The question is:
Prove that if $U$ is open in $\mathbb{R}^n$ and $A$ is an open subset of $U$ then $\overline{A}\subseteq U$.
I want to prove that $\overline{A}\subseteq U$ by contradiction.
I thought of, say, the open set
$$ A = [0,1)\cup(2,3)$$
but $A$ is not open and so I can’t use the definition of $\overline{A}$ as the largest open subset of $A$, because $A$ is not open.
Can anyone help?
A:
1) Consider the complement $U^c=\mathbb{R}^n \setminus U$ of $U$.
2) Observe that $A$ is an open subset of $U$, since it is the union of open sets.
Therefore $\mathbb{R}^n\setminus A$ is closed, since it is the union of two closed sets. So, $\overline{A}\subset U$
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