$(a)$ Let $S$ be a non-empty set (finite or infinite), and consider the space $X = \mathscr{B}(S)$ of all bounded, real-valued functions on $S$. Define $d_{\infty}$ on S by:
$d_{\infty}(f, g) = \sup_{s\in S}|f(s) − g(s)|$
which is a metric. Show that $(X, d_{\infty})$ is complete.
$(b)$ Consider $\mathcal{C}[−1, 1]$, the space of continuous functions, as a subset of $X$. Is $\mathcal{C}[−1, 1]$ closed in $X$?
For (a) i know f is a lipschitz mapping if that helps.
To show that $(X, d_{\infty})$ is complete, pick a Cauchy sequence $\{ f_n\}_n \subseteq X$. This means that $$\forall \varepsilon >0, \exists N : \forall n,m \geq N \ \ \ \sup_{s \in S} |f_n(s) - f_m(s)| < \varepsilon$$
In particular this implies that $\forall s \in S$ the sequence $\{ f_n(s) \}_n \subseteq \mathbb{R}$ is Cauchy. But $\mathbb{R}$ is complete, so $\{ f_n(s) \}_n$ is convergent. Call $$f(s) = \lim_n f_n(s)$$
In this way you define the pointwise limit of the $f_n$. Now you have to show two things:
This completes the proof, since you show that every Cauchy sequence is convergent.
Since $f_n$ is Cauchy, it is a bounded sequence. This means that $\exists L >0$ such that for all $n$ you have $\sup_{s \in S} |f_n(s) - f_1(s)| \leq L$. But now, this implies that for all $s \in S$ you have $|f(s)| \leq L + \sup_S |f_1|$, so $f$ is bounded.
$\forall \varepsilon > 0$ let $N$ be such that $\forall n,m \geq N \ \ \sup_{s \in S} |f_n(s) - f_m(s)| < \varepsilon$. Then $\forall n \geq N$
$$d_{\infty} (f, f_n) = $$
