If $U\subseteq X'$ is open relative to $X'$, and $X'\subseteq X$ is open relative to $X$ Here $X'$ is a topological subspace of $X$. If the above conditions hold, then is it true that $U$ is open in $X$? This is a part of a proof which I don't quite fully understand.
 A: Yes! Open in open is open. Since $U$ is open in $X'$, then there exists an open subset $V$ of $X$ such that $U=X'\cap V$. The fact that $X'$ is open in $X$ gives that $U$ is a finite intersection of open subsets of $X$. Thus $U$ is open.
Along the same lines, closed in closed is closed.
A: No proof is needed because it is just the definition of the subspace topology, that is in the subset topology of $X'\subseteq X$, $X$ topological space, the open sets are all and only the intersections of every single open set of $X$ with $X'$
$$\mathcal{T}_X(X')=\{G'\subseteq X'|~\exists G\in\mathcal{T}(X),G'=G\cap X'\}$$
where $\mathcal{T}(X)$ is the topology of $X$ and $\mathcal{T}_X(X')$ is the subset topology of $X'$ induced by $\mathcal{T}(X)$.
What's need to be proved, and probably what you were searching for, is the fact that such a family of set $\mathcal{T}_X(X')$ is indeed a topology.
Proof of the closure of $X'$ w.r.t. the union of open sets.
$\forall \mathcal{G}'\in\mathcal{T}_X(X'),~\cup \mathcal{G}'=\cup\{G\cap X'|~\exists G'\in\mathcal{G}',~G'=G\cap X'\}=(\cup\mathcal{G})\cap X'\in\mathcal{T}(X')$ 
where $\mathcal{G}=\{G\in\mathcal{T}(X)|~\exists G'\in\mathcal{G}',~G'=G\cap X'\}$. The last equality holds true because $\cup\mathcal{G}\in\mathcal{T}(X)$ for the closure of $X$ w.r.t. the union of open sets, and the definition of open sets in $X'$.
Proof of the closure of $X'$ w.r.t. the intersection of finitely many open sets.
$\forall G'_1,G'_2\in\mathcal{T}_X(X'),~G'_1\cap G'_2=G_1\cap G_2\cap X'\in\mathcal{T}_X(X')$
where $G_i\in\mathcal{T}(X)$ such that $G'_i=G_i\cap X', i=1,2$. The last equality holds true because $G_1\cap G_2\in\mathcal{T}(X)$ for the closure of $X$ 
w.r.t. the intersection of finitely many open sets and the definition of open sets in $X'$.
