Separation properties of metric spaces

Prove the stated assertion. In any normed space there exist two disjoint open sets that are not contained in any disjoint closed sets.

Prove the stated assertion. A metric space is a Hausdorff topological space if its topology is understood as the collection of all its open sets, defined by means of the metric.

Let $X$ be a compact metric space, let $f \in C(X)$, and let $\varepsilon >0$. Suppose that $\left\{ F_{k}\right\}_{k=1}^{n}$ are disjoint closed subsets of $X$. Prove that there exist disjoint open sets $\left\{ G_{k}\right\}_{k=1}^{n}$ such that, for all $1 \leq k \leq n, F_{k} \subset G_{k}$ and $\inf \left\{ f(x): x \in G_{k}\right\} \geq \inf \left\{ f(x): x \in F_{k}\right\} -\varepsilon$.

Let $A$ and $B$ be disjoint nonempty closed sets in a metric space $X$, and define

Show that $f$ is a continuous function on $X$ whose range lies in $[0,1]$, that $f(p) = 0$ precisely on $A$ and $f(p) = 1$ precisely on $B$. This establishes a converse of Exercise 3: Every closed set $A \subset X$ is $Z(f)$ for some continuous real $f$ on $X$. Setting

show that $V$ and $W$ are open and disjoint, and that $A \subset V, B \subset W$. (Thus pairs of dis joint closed sets in a metric space can be covered by pairs of disjoint open sets. This property of metric spaces is called normality).