Space of bounded functions is complete

Let $(f_n)_{n \in \Bbb N}$ be a Cauchy sequence in $B(T,X)$. We have to show it has a limit.

For each $t \in T$ it's clear that $d(f_n(t), f_m(t)) \le d_\infty(f_n, f_m)$ as the latter is defined as te supremum (so an upperbound) for all such distances. It immediately follows from the definitions that $(f_n(t))_n$ is a Cauchy sequence in $(X,d)$. And as $(X,d)$ is complete it has a unique limit in $X$, which I call $f(t)$ (as it depends on $t$). This defines a function $f: T \to X$. Note that by definition $\lim_{n \to \infty} f_n(t) = f(t)$ for all $t$.

This is our candidate limit for $(f_n)_n$. Is it bounded? Let $\varepsilon=1$ in the definition of Cauchy and we get $N \in \Bbb N$ such that for all $n,m \ge N$ we get $d_\infty(f_n, f_m) < 1$.

Then $f_N \in B(X,T)$ so there is some $R >0$ such that $f_N[T] \subseteq B(x_0,R)$ for some $x_0 \in X$. Now for $t \in T$ fix $N_1$ so that $n \ge N_1$ implies $d(f_n(t), f(t)) < 1$ as well, and then $$d(f(t), x_0) \le d(f_n(t), f(t)) + d(f_n(t), f_N(t)) + d(f_N(t), x_0) < 1 + 1 + R$$

for all $n \ge \max(N, N_1)$ and it follows that $f[T] \subseteq B(x_0, R+2)$ so that $f$ is indeed bounded.

It suffices to show that $f_n \to f$ under $d_\infty$. So let $\varepsilon>0$ be given. Let $N$ be the index given by the Cauchy-ness of $(f_n)_n$ for $\frac{\varepsilon}{2}$. So for all $n,m \ge N$ we have, for any fixed $t$:

$$d(f_n(t), f_m(t)) \le d_\infty(f_n,f_m) < \frac{\varepsilon}{2}$$

Now we can use that $d$ is a continuous function on $(X,d)$ and let $m \to \infty$ so that $d(f_n(t), f(t)) \le \frac{\varepsilon}{2}$ and as this holds for all $n \ge N$ and all $t$ we have that for all $n \ge N$: $$d_\infty(f_n, f) < \varepsilon$$ showing the required convergence.