Let $X$ be the set of all binary sequences $f:\mathbb{N} \rightarrow \{0,1\}$. Define the metric $d$ on $X$ by $$ d(f,g)= \begin{cases} 0&f=g\\ \frac{1}{2^m}&m=\min\{n|f(n)\neq g(n)\}\\ \end{cases} $$ a) Prove that $(X,d)$ is compact.
b) Prove that no point in $(X,d)$ is isolated.
My idea: For part a), we need to show that it is sequentially compact.i.e every sequence in $X$ has a convergent subsequence.
Let $g_n$ be any sequence in $X$., then construct a sequence,
$g_1(1),g_2(1),........$, there are infinitely many terms of the sequence either goes to $0$ or $1$.
Similarly,$g_1(2),g_2(2),........$, there are infinitely many terms of the sequence either goes to $0$ or $1$.
Continue this way, there are infinitely many sequence that goes to $1$ or $0$.
Can anyone suggest me , how I move further to construct a subsequence?
$X$ can be represented in an equivalent way as $\{0,1\}^{\mathbb{N}}=\{(x_n)_{n\in \mathbb{N}} : \quad x_n \in \{ 0,1\}, \forall n\in \mathbb{N} \}$. Observe that the metric $d$ on $X \equiv \{0,1\}^{\mathbb{N}}$ is actually compatible with the product topology when $\{0,1\}$ is endowed with the discrete topology. In particular, $\{0,1\}$ (with the discrete topology) is compact. By Tychonoff's Theorem, we have that $(X,d)$ is compact.