Let all functions $x: [0,1] \rightarrow \mathbb{R}$ where each of function $x$, we have points $\{t_k: t_k \in [0,1], k \in \mathbb{N}\}$ such that $x(t_k) \neq 0$ $\forall k \in \mathbb{N}$ and $\sum_{k=1}^{\infty} x^2(t_k) < \infty$. And we define in this space addition and multiplication operations as usual and dot product by
$(x,y)=\sum_{k=1}^{\infty}x(t_k)y(t_k)$ such that $x(t_k)y(t_k) \neq 0$.
Problem. There is no always dense countable subset.
In this problem, I supposed that exists dense countable subset. Let $\{x_n\}_{n \in \mathbb{N}}$ a dense countable subset, then exists for all $n \in \mathbb{N}$, $\{t_{k_n}\}_{k \in \mathbb{N}}$ such that $x(t_{k_n})\neq 0$ and $\sum_{k=1}^{\infty} x^2(t_{k_n}) < \infty$. But I don't how to proceed, how can I use that $\{x_n\}_{n \in \mathbb{N}}$ is dense. Could you help me to finish this problem? please.
Define $x_t(s)=1$ for $s=t$ and $0$ for $s \neq t$. Verify that $\|x_t-x_u\|=\sqrt 2$ whenever $t \neq u$. Thus there are uncountably many vectors at distance $\sqrt 2$ from each other. This implies that the space is not separable.
If $D$ is a coutable dense set we can pick an element of $D$ from each of the balls $B(x_t,\frac 1 {\sqrt 2})$. This gives a ont-to-one map from $[0,1]$ into $D$, making $D$ necessarily uncountable.