Question: Let $X$ be a regular space, then $\omega(X)$ $\le$ $2^{\delta(X)}$, where $\delta(X)$ is the density of $X$. I don't know how to prove this, or how to apply the hint that $\mathscr B = \{\text{int}(\text{cl}(V)):V\subset D\}$ is a basis for $X$, and $|D|=\delta(X)$.

Answer: 
I'll use $w(X)$ for weight and $d(X)$ for density, as is more common.

The [regular open sets](https://en.wikipedia.org/wiki/Glossary_of_topology#R) $\mathrm{RO}(X)$ form a base for a regular $X$: let $O$ be open and $x \in O$. Then regularity gives us an open set $V$ such that $$x \in V \subseteq \overline{V} \subseteq O$$

and $\operatorname{int}(\overline{V})$ is then a regular open set that contains $x$ and sits inside $O$.

Also, by standard facts (and not too hard to show): $\operatorname{int}(\overline{V}) =\operatorname{int}(\overline{V \cap D})$ when $D$ is a dense subset of $X$. (Even $\overline{V \cap D} = \overline{V}$ holds, in all spaces.)

Now $w(X) \le |\mathrm{RO}(X)|$ is clear as the second is the size of *one* base for $X$ and $w(X)$ the minimal size of a base.

And the closure fact in the standard paragraph implies, that taking a dense subset $D$ of $X$ of minimal size $d(X)$, the map 

$$\mathrm{RO}(X) \ni O \mapsto (O \cap D) \in \mathscr{P}(D)$$ is one-to-one
showing that $|\mathrm{RO}(X)| \le 2^{d(X)}$ and then combine these facts.
Let $P(D)$ be the power-set of $D,$ the set of all subsets of $D.$ 

Since $\mathscr B$ is a base (basis) for $X$ we have $$w(X)\le |\mathscr B|+\aleph\_0=$$ $$=|\{int (cl (V)):V\in P(D)\}|+\aleph\_0\le$$ $$\le |\{V: V\in P(D)\}|+\aleph\_0=$$ $$=|P(D)|+\aleph\_0=$$ $$=2^{|D|}+\aleph\_0=$$ $$=2^{\delta(X)}+\aleph\_0=2^{\delta}.$$