Justifying that "metric space of bounded functions" fails the second axiom of countability? Self studying topology from Kahn, an exercise asks to find a metric space that fails the second axiom of countability. I can't find the name of the metric space that I thought of online, and when I search the quoted phrase it comes up with papers much more advanced than what I am trying to figure out. I'd like to self-evaluate. This is my solution. Any critique is welcomed. 
Let $F$ be the set of all bounded functions from $[0,1]$ into $\Bbb R$ under the metric 
$$D(f_1, f_2) = \max\{|f_1(x) – f_2(x)|: x \in [0,1]\}\;.$$
Consider a countable subset of the basis $\{B_\epsilon(f) \mid f \in F, \epsilon\in\Bbb R\}$. For each $k \in\Bbb N$ we have an open set $U_k$ of the form $B_\epsilon(f)$. For each $\alpha\in [0,1]$ let $U_k^\alpha$ denote the interval $(f(α) – \epsilon, f(α) + \epsilon)$. 
I wish to show that there is a function $f \in F$ which is not contained by any $U_k$, and thus no countable basis can exist. 
Define $g:[0,1]\to\Bbb R$ as 
$$g(x)=\begin{cases}
1+\sup U_k^x,&\text{if }x=\frac1k\text{ for some }k\in\Bbb N\\
0,&\text{otherwise}\;.
\end{cases}$$
Thus $g$ is not contained in any $U_k$ since it is outside of the epsilon radius for at least one $\alpha$ in each $U_k$. 
Therefore there is no countable basis for this metric space. 
Again please critique and suggest better notation for clarity. Posting from my phone and cannot access the latex tips.
 A: It would be better to name the function and $\epsilon$ used to define each $U_k$: for each $k\in\Bbb Z^+$ there are $f_k\in F$ and $\epsilon_k>0$ such that $U_k=B_{\epsilon_k}(f_k)$. Then you can define $g$ by
$$g(x)=\begin{cases}
1+f_k(x)+\epsilon_k,&\text{if }x=\frac1k\text{ for some }k\in\Bbb Z^+\\
0,&\text{otherwise}
\end{cases}$$
and don’t even need to define the intervals $U_\epsilon^x$. In fact you don’t even need the $1$:
$$g(x)=\begin{cases}
f_k(x)+\epsilon_k,&\text{if }x=\frac1k\text{ for some }k\in\Bbb Z^+\\
0,&\text{otherwise}
\end{cases}$$
works just as well, since
$$D(g,f_k)\ge\left|f_k\left(\frac1k\right)-f_k\left(\frac1k\right)\right|=\epsilon_k$$
for each $k\in\Bbb Z^+$.
Your argument requires one non-trivial fact that you’ve not mentioned: it depends on the fact that if $\mathscr{B}$ is a base for a second countable space, then there is a countable $\mathscr{B}_0\subseteq\mathscr{B}$ that is countable. That is, if a space is second countable, then every base for the space contains a countable base for it. This is the countable case of a more general theorem that I stated and proved in this answer.
