Prove a metric space is totally bounded

If I'm understanding the question correctly, what you're looking for is, for each $varepsilon>0,$ a finite set of points in your space such that every point in the space is within a distance $varepsilon$ of some point in that finite set.

Find the smallest positive integer $k$ such that $2^{-k}<varepsilon.$ Then consider the set of all sequences of the following form: $$x_1, x_2, x_3, ldots,x_k,,underbrace{0, 0, 0, 0, 0, ldotsldots}$$ There are only $2^k$ of these, a finite number. And every point is within $varepsilon$ of one of these.

It is enough to consider the special case of an $epsilon$ of the shape: $$ epsilon = frac 1{2^{N-1}} ,qquad NinBbb Z_{>0} . $$ So let us fix such an $N$ and a corresponding $epsilon$. Consider all (finitely many) elements $x(a)=(a_1,a_2,dots,a_N,0,0,dots)$ with $a=(a_1,a_2,dots,a_N)in{0,1}^{times N}$ that have all possible starts on the first $N$ places, followed by zeros.

Consider now an $x$ in the metric space, select the $x(a)$ with the match with $x$ on the first $N$ places and estimate the distance from $x$ to $x(a)$ by being generous on the places on positions $>N$.