I'm trying to prove that the space of bounded functions $B\left([0,1]\right)$ is not separable. My idea is first construct a "simple" uncountable—and bounded—sequence of functions on the interval $[0,1]$. So for $\alpha \in [0,1]\setminus \mathbb{Q}$, I define $f_{\alpha} \in B\left([0,1]\right)$ by
$$f_{\alpha}(x) = \left\{ \begin{array} 11, \, x = \alpha \\0, \text{ otherwise} \end{array} \right.$$
Is this construction valid? Aren't sequences indexed by the naturals? Maybe I should call this a family of functions...
Anyways, from here, I suppose that $M$ is dense in $B([0,1])$. Then, for any ball $V_{\alpha}$ around a particular $f_{\alpha}$, that ball must contain an element of $M$. By the way the metric I'm using is
$$d(x,y) = \sup_{t\in [0,1]}|x(t) - y(t)|$$
Considering $V_\alpha$ with radius $1/3$ we have
$$V_{\alpha}\left(f_{\alpha}; \frac{1}{3}\right) = \left\{ f \in B([0,1]) : d(f,f_{\alpha}) < \frac{1}{3} \right\}$$
so $V_\alpha \cap M \neq \emptyset$, but $V_\alpha \cap V_\beta = \emptyset$ since $d(f_\alpha, f_\beta) = 1$, for the functions described above.
Therefore, I conclude that any dense set $M$ in $B([0,1])$ is uncountable, since it has an uncountable subset*.
*Specifically the uncountable subset is
$$\left(V_\alpha \cap M\right) \cup \left(V_\beta \cap M\right) \cup \cdots$$
Just looking for guidance/verification on the proof, not an answer necessarily. Thank you.
$(f_{\alpha})$ is a family of functions, not a sequence. You can pick an element $g_{\alpha}$ from $V_{\alpha}(f_{\alpha},\frac 1 3)$ for each $\alpha$ and say that the map $\alpha \to g_{\alpha}$ is a one-to-one map from the uncountable set $[0,1]\setminus \mathbb Q$ into the countable set $M$ which is a contradiction. This argument brings in more clarity.
And there is no reason to remove $\mathbb Q$ from the index set.