Closed Set in Topological Subspace

Theorem
Let $T$ be a topological space.

Let $T' \subseteq T$ be a subspace of $T$.

Then $V \subseteq T'$ is closed in $T'$ $V = T' \cap W$ for some $W$ closed in $T$.

Necessary Condition
Suppose $V \subseteq T'$ is closed in $T'$.

Then $T' \setminus V$ is open in $T'$ by definition.

So, by definition of subspace topology, $T' \setminus V = T' \cap U$ for some $U$ open in $T$.

Then:

Thus $T \setminus U$ is closed in $T$.

Sufficient Condition
Conversely, suppose $V = T' \cap W$ where $W$ closed in $T$.

Then $T' \setminus V = T' \setminus \left({T' \cap W}\right) = T' \cap \left({T \setminus W}\right)$ which is open in $T'$.

So $V$ is closed in $T'$.

Also see

 * Neighborhood in Topological Subspace iff Intersection of Neighborhood and Subspace