Proposition

Suppose $S$ is a subspace of the topological space $X$.

If $U \subset S \subset X$, $U$ is open in $S$, and $S$ is open in $X$, then $U$ is open in $X$. The same is true with “closed” in place of “open.”
If $U$ is a subset of $S$ that is either open or closed in $X$, then it is also open or closed in $S$, respectively.

Solution

1

If $U$ is open in $S$, then $U = S \cap V$ for some $V \subset X$ that is open in $X$ by the definition of a subspace topology. Since the intersection of two open sets is open, $U = S \cap V$ is open in $X$.

Suppose that $C$ is closed in $S$ and $S$ is closed in $X$. Then $S \setminus C$ is open in $S$. Therefore, $S \setminus C = S \cap U$ for some set $U \subset X$ that is open in $X$. Then $C = S \setminus (S \setminus C) = S \setminus (S \cap U) = S \setminus U = S \cap (X \setminus U)$. Then $S$ is a closed subset of $X$ and $X \setminus U$ is closed in $X$. Since the intersection of closed subsets is closed, $C = S \cap (X \setminus U)$ is closed in $X$.

2

Let $U \subset S$. Suppose $U$ is open in $X$. Then $U = S \cap U$, so $U$ is open in $S$.

Let $C \subset S$. Suppose $C$ is closed in $X$. Then $X \setminus C$ is open in $X$. Thus $S \cap (X \setminus C)$ is open in $S$. $S \cap (X \setminus C) = S \setminus C$ because $S \subset X$. Then $S \setminus C$ is open in $S$. Thus $C$ is closed in $S$.