Problem 1:
Every metric space is normal.
My attempt: Let $X$ be a metric space. Let $A,B$ be disjoint, closed subsets of $X$. Then, since $A\subseteq X\backslash B$, for each $x\in A$, there exists $r_x>0$,for which, $B(x,r_x) \cap B=\varnothing$. Let $U=\bigcup_{x\in A}B(x,\frac{r_x}{2})$ and $V= \bigcup_{y\in B}B(y,\frac{r_x}{2})$. Then, I claim: $U\cap V=\varnothing$. Suppose otherwise, so if they share a common element, $z$. Then, for some $x\in A$ and some $y\in B$, $d(x,y)\leq d(x,z)+d(y,z)<r_x$. This would imply that $y\in B\cap B(x,r_x)$. Contradiction.
Problem 2: Every closed subspace of a normal space is normal.
My attempt: Let $X$ be a normal topological space. Suppose $F\subseteq X$ is closed. Let $A,B$ be disjoint closed subsets of $F$. So, $A=A'\cap F$ and $B=B'\cap F$, where $A', B'$ are closed in $X$ that are disjoint. Since $X$ is normal, there exists open subsets (in X), $U_1$ of $A'$ and $U_2$ of $B'$, such that $U_1\cap U_2=\varnothing$. Hence, $U_1\cap F \supseteq A$ and $U_2\cap F\supseteq B$, both of which are closed in $F$.
Are my attempts correct?
The first needs a little more attention to detail (e.g. two open sets intersect and you're talking about one $r_x$?)
For the second just note that $A \subseteq F$ closed in $F$ is itself already closed in $X$ too! So separate $A,B$ in $X$ and intersect the open sets with $F$.