Let $X$ be the space of sequences $x = (x_1, x_2, ..., x_n,...)$ such that $x_i = 0$ or $x_i=1$, equipped with the metric $$d(x,y) = \sum_{n=1}^{\infty} \frac{1}{2^n} \left| x_n - y_n \right|.$$ Prove that $(X,d)$ is both complete and compact.
Attempt: Suppose $(x_n)$ is a Cauchy sequence in $(X,d)$. Then $\forall \epsilon > 0 \ \exists N \in \mathbb{N}$ such that $$m,n > N \implies \sum_{n=1}^{\infty} \frac{1}{2^n} \left| x_n - x_m \right| < \epsilon.$$ Consider that \begin{eqnarray*} \sum_{n=1}^{\infty} \frac{1}{2^n} \left| x_n - x_m \right| & = & \sum_{n=1}^{\infty} \frac{1}{2^n} \left| x_n - x + x - x_m \right| \\ & \leq & \sum_{n=1}^{\infty} \frac{1}{2^n} \left| x_n - x \right| + \sum_{n=1}^{\infty} \frac{1}{2^n}\left| x-x_m \right| \\ & \leq & \frac{1}{2^n} + \frac{1}{2^n} \\ &=& \frac{2}{2^n} = \frac{1}{2^{n-1}} \to 0 \ \text{as} \ n \to \infty. \end{eqnarray*} Therefore, $x_n \to x$, and $(X,d)$ is a complete metric space.
How do we then show that $X$ is compact with respect to this metric?
After showing completeness it suffices to prove that $X$ is totally bounded.
That is for every $\epsilon$ $X$ can be written as a union of finitely many balls.
Choose an $\epsilon >0$ then pick $M$ such that $\sum_{n\geq M+1} \dfrac{1}{2^n} < \epsilon$.
Now the points which will be the centers of the balls are of the form $x_i=\{ (u_1,u_2, \cdots, u_M,0,\cdots,0\cdots )\}$, where $u_i=0$ or $1$ so you get $2^M$ many points.
Now I leave it as a exercise to show that indeed $X= \bigcup_{i=1}^{2^M}B(x_i,\epsilon)$ where $B(x_i,\epsilon)= \{x \in X | d(x_i,x)<\epsilon \}.$