I have almost solved the following problem but am stuck at the very end, can you help me finish it? Thank you for your help.
Let $n<\omega$ and $t\in {}^n\omega$. We define $U_t=\{s\in {}^\omega\omega : t\subseteq s\}$. The family $\mathcal B=\{U_t : t\in\bigcup {}^n\omega\}$ is a basis for a topology $\tau$ on ${}^\omega\omega$. This means that a set $X$ is in $\tau$ if and only if $X$ is a union of elements of $\mathcal B$.
EXERCISE 50 (PG): Show that $\langle {}^\omega\omega,\tau \rangle$ is a second countable completely metrizable space. Hint: For $r,s\in {}^\omega\omega$ such that $r\ne s$, the $\varrho(r,s)=1/\min\{n : r(n)\ne s(n)\}$. Moreover, let $\varrho(r,r)=0$. Show that $\varrho$ is a complete metric on ${}^\omega\omega$ that induces the topology $\tau$.
(i) Second countable: the set of $n$-tuples is countable since it is a finite union of countable sets.
(ii) complete with respect to $ d(f,g) = \frac{1}{\min \{n \mid f(n) \neq g(n) \}}$ Let $f_k$ be a Cauchy sequence. This means that the index at which $f_k, f_i$ differ gets larger as $i,k$ get larger. Let $f(n)$ denote the pointwise limit, that is, $\lim_{k \to \infty} f_k(n)$. Then $f(n) \in \omega$ for every $n$ since $f_k(n)$ is eventually constant. Hence $f \in \omega^\omega$.
(iii) $d$ induces the same topolgy as $U_t$:
$\tau_t \subset \tau_d$: Let $s \in U_t$. Then $B(s, \frac{1}{n+1})$ is an open ball contained in $U_t$ hence $U_t$ is open in $\tau_d$.
$\tau_t \supset \tau_d$: Let $B(s, \frac{1}{k})$ be an open balls. If $k < n$, $s \in U_s \subset B(s, \frac{1}{k})$. But if $k \ge n$ then I'm stuck. The ball $B(s, \frac{1}{k})$ consists of all sequences that agree with $s$ on the first $k$ coordinates. So it is quite small. How do I make the $U_t$ small enough to fit in this ball? I have thought of intersection but since I'm only allowed to do finite intersections I don't see what to do.