We know that a metric space $(X,d)$ is said to be totally bounded if for any $\varepsilon > 0$, there are finitely many points $\{x_{1}, \dots, x_{k_{\varepsilon}}\}$ such that $X \subseteq \bigcup_{n=1}^{k_{\varepsilon}} B_{\varepsilon}(x_{n})$.
Now, what if we were to alter this definition to change finitely many points $x_{i}$ to countably many points $x_{i}$? When we prove that every totally bounded space is separable (i.e. we can find a countable dense subset), we never really use the fact that total boundedness gives us finite covers of every possible radius, but only that we get countable covers of every possible radius. It is, however, possible that I'm missing something. My question: is it possible to get separability from the weaker notion of total boundedness that I have described above?
Yes, a space which satisfies your weaker condition is separable.
For every $n$ let $\{B^i_n\}_{i\in\Bbb{N}}$ be countably many balls of radius $1/n$ covering $X$ and let $\{q^i_n\}_{i\in\Bbb N}$ be their centers. I claim that $$D=\bigcup_{n\in\Bbb N}\bigcup_{i\in\Bbb N}\{q^i\}$$ is a countable dense set. Let $x\in X$ be arbitrary and let $\varepsilon>0$, we want to find a $y\in D$ with $d(x,y)<\varepsilon$. Let $N$ be big enough such that $1/N<\varepsilon$, by hypothesis there is a ball of radius $1/N$ containing $x$, so its center is the $y$ we're looking for.