Prove that the space of sequences under this metric is complete and compact. I'm currently studying for the prelim exams, and I would love a hint on how to complete this problem.

If $X$ is the space of sequences of $0$'s and $1$'s (i.e., $x \in X$ if $x = (x_{1}, x_{2}, x_{3}, \dots )$ with $x_{i} = 0$ or $1$), and $d(x,y) = \sum \limits_{n = 1}^{\infty} \dfrac{|x_{n} - y_{n}|}{2^{n}}$, then prove that $d$ is complete and $(X, d)$ is compact.

I'm not even sure how to start this.  All that I came up with is the seemingly obvious fact that $d(x,y) \leq 1$ for any $x, y \in X$.
I know to show the metric is complete, I need to show that any Cauchy sequence converges (or has a convergent subsequence...).
So, let $\{x_{n}\}_{n = 1}^{\infty} \subseteq X$ be a Cauchy sequence.  Then for every $\epsilon > 0$, $\exists N \in \mathbb{N}$ such that $m, n \geq N$ implies $d(x_{n}, x_{m}) < \epsilon$.  Thus, $\sum \limits_{k = 1}^{\infty} \dfrac{|x_{m_{k}} - x_{n_{k}}|}{2^{k}} < \epsilon$.
Now what?  I have some sort of idea that as $m$ and $n$ get bigger, the starting terms of each sequence must be equal...I think that must be true.  For example, if $\epsilon < \frac{1}{2}$, then certainly we would need $x_{n_{1}} = x_{m_{1}}$.  Similarly, if $\epsilon  < \frac{1}{4}$, then we would need $x_{n_{2}} = x_{m_{2}}$ as well....

By the Tychonov theorem, this space is compact. It is also metrizable via the metric
$$\rho(x,y) = \sum_{n=1}^\infty {|x_n - y_n|\over 2^n}.$$
A compact metrizable space is complete.
