I would like to prove that the space $\{-1,1\}^\mathbb{N}$ is compact under the metric $$d(s_1, s_2)=2^{-\inf\{k\in\mathbb{N}: \,s_1(k)\neq s_2(k)\}}. $$ Since this is a metric space, compactness is equivalent to sequential compactness. Thus, given a sequence $\{s_j\}_{j\ge 1}=(s_j^{(1)}, s_j^{(2)}, \cdots)\subseteq\{-1, 1\}^\mathbb{N}$, I need to find a subsequence that converges with respect to $d$. I think I am supposed to use a kind of diagonal argument.
More specifically, I need to find an element $t\in\{-1, 1\}^{\mathbb{N}}$ such that given any $k\ge 1$, I can find an element of the sequence, $s_{n_k}$, such that $s$ and $s_{n_k}$ agree in their first $k$ digits. Then we would have $d(s, s_{n_k})< 2^{-k}$ and $s_{n_k}$ would converge to $s$.
Let us write the sequences as follows \begin{aligned} s_1&=(s_1^{(1)}, s_1^{(2)}, \cdots)\\ s_2&=(s_2^{(1)}, s_2^{(2)}, \cdots) \\ &\vdots\\ s_j&=(s_j^{(1)}, s_j^{(2)}, \cdots), \end{aligned} where $s_j^{(k)}\in\{-1,1\}$ for all $(k, j)\in\mathbb{N}^2$.
My idea: If we look at the first "vertical sequence" $s^{(1)}_j$ for $j\ge 1$, there exists an infinity of at least one of $1$ or $-1$. Assume it is $1$. Then, I can define $t^{(1)}=1$. Similarly, define $t^{(m)}$ as either $-1$ or $1$ chosen such that the "vertical" sequence $s_j^{(m)}, j\ge 1$ contains an infinite number of the chosen number. Then, I would like to say that given any $k$, we can find an element $s_{n_k}$ such that $t$ and $s_{n_k}$ agree up to the first $k$ elements. Is this the right approach?