Metric space of infinite binary sequences Let $\Omega = \{0,1\}^{\mathbb{N}}$ be the space of infinite binary sequences. Define a metric on $\Omega$ by setting $d(x,y) = 2^{-n(x,y)}$ where $n(x,y)$ is defined to be the maximum $n$ such that $x_i = y_i$ for all $i\le n$. Show that $(\Omega, d)$ is a compact metric space.
I have tried to show this by taking open covers of $\Omega$ and finding a finite subcover, but that does not seem to be working. How could I approach this problem?

HINT: Give $\{0,1\}$ the discrete topology, and let $\tau$ be the resulting product topology on $\Omega$; $\Omega$ is certainly compact in this topology, since it’s a product of compact spaces. If $\tau_d$ is the topology generated by the metric $d$, show that $\tau_d=\tau$.
Alternatively, let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a sequence in $\Omega$, where $x_n=\langle x_n(k):k\in\Bbb N\rangle$, and show that $\sigma$ has a convergent subsequence. To do this, note first that there must be a $b_0\in\{0,1\}$ and an infinite $N_0\subseteq\Bbb N$ such that $x_n(0)=b_0$ for each $n\in N_0$. Then there must be a $b_1\in\{0,1\}$ and an infinite $N_1\subseteq N_0$ such that $x_n(1)=b_1$ for each $n\in N_1$. Keep going.
