Definitions:
A set $A$ in a metric space $X$ is said to be an $\epsilon$-net in $Y$ if for each $y\in Y,\exists a\in A$ such that $d(a,y)<\epsilon$.A set $A$ is said to be an universal net of $Y$ if $A$ is an $\epsilon$-net of $Y$ for each $\epsilon>0$.
Lemma:
$D$ is dense in $X$ iff $D$ is an universal net of $X$.
Proof: (Only if) $D$ is dense in $X$.Now take any $\epsilon>0$,then consider any $x\in X$,$B(x,\epsilon)\cap D\neq \phi$,so there exists $z\in D$ such that $d(x,z)<\epsilon$.So,$D$ is an $\epsilon$-net of $X$,but $\epsilon$ is arbitrary.So,$D$ is a universal net of $X$.
(If)Conversely,let $D$ be an universal net of $X$,then take any $x\in X$ and consider any $\epsilon>0$.Now,$D$ is an $\epsilon$ net of $X$,so $d(x,z)<\epsilon$,for some $z\in D\implies z\in B(x,\epsilon)\cap D$.So,$D$ is dense in $X$.
Now we move to our next claim:
Theorem: If $X$ is totally bounded,then $X$ is separable.
Proof: $X$ is totally bounded,so $X$ has a finite $\epsilon$-net for each $\epsilon$.Take the finite $1/n$-nets and consider their union $D$.Then this set $D$ is an universal net of $X$ and hence dense in $X$,also $D$ is a countable union of finite sets and hence countable.So,$X$ is separable.
This proof is quite simple,I think this is a good notion because total boundedness means existence of finite nets of every radius and density means existence of a universal net.Thinking this way makes it more intuitive and visually clear.