Open Set in Open Subspace

Theorem
Let $X$ be a topological space.

Let $U\subset X$ be an open subset.

Let $V\subset U$ be a subset.

Then $V$ is open in $U$ $V$ is open in $X$.

Proof
Let $V$ be open in $X$.

By Intersection with Subset is Subset, $V\cap U = V$.

By definition of topological subspace, $V$ is open in $U$.

Let $V$ be open in $U$.

By definition of topological subspace, there exists an open subset $W\subset X$ with $V=U\cap W$.

Because $U$ and $W$ are open in $X$, $V=U\cap W$ is open in $X$.

Also see

 * Closed Set in Closed Subspace