I want to cross-check one fact:
Suppose I have $$P_{12}=[0,1]\times \{1\} \times \{0\} \times \cdots=\{(x_n)| x_1\in [0,1],x_2=1,x_i=0, \text{ for }i\neq 1,2\}$$ $$P_{13}=[0,1]\times \{0\} \times \{1\} \times \cdots=\{(x_n)| x_1\in [0,1],x_3=1,x_i=0, \text{ for }i\neq 1,3\}$$ $$\vdots$$ $$P_{1n}=[0,1]\times \{0\} \times \{0\} \times \cdots \{1\}(\text{ nth position}) \times \{0\} \times \cdots =\{(x_n)| x_1\in [0,1],x_n=1,x_i=0, \text{ for }i\neq 1,n\}$$ $$\vdots$$ $$P_{23}=[0,1]\times [0,1]\times \{1\} \times \{0\} \times \cdots=\{(x_n)| x_1,x_2\in [0,1],x_3=1,x_i=0, \text{ for }i\neq 1,2,3\}$$ $$\vdots$$ $$P_{2n}=[0,1]\times [0,1]\times \{0\} \times \{0\} \times \cdots \{1\}(\text{ nth position}) \times \{0\} \times \cdots =\{(x_n)| x_1,x_2\in [0,1],x_n=1,x_i=0, \text{ for }i\neq 1,2,n\}$$ Similarly we define $$P_{nm}=[0,1]\times [0,1]\times \cdots \times [0,1](\text{ n times}) \times \{0\} \times \{0\} \times \cdots \{1\}(\text{ mth position}) \times \{0\} \times \cdots=\{(x_n)| x_1,x_2,\cdots,x_n\in [0,1],x_m=1,x_i=0, \text{ for }i\neq 1,2,\cdots,n,m\}$$ where $m>n$.
Now let $S=\bigcup_{i<j=1}^{\infty}P_{ij}$. Is $S$ separable? I think it is. I know that $\Bbb Q^{\infty}$ is not countable but $$Q_{1n}=([0,1]\cap \Bbb Q)\times \{0\} \times \{0\} \times \cdots \{1\}(\text{ nth position}) \times \{0\} \times \cdots =\{(x_n)| x_1\in ([0,1] \cap \Bbb Q),x_n=1,x_i=0, \text{ for }i\neq 1,n\}$$ is countable $\forall n$. Similarly, $$Q_{2n}=([0,1]\cap \Bbb Q)\times ([0,1]\cap \Bbb Q)\times \{0\} \times \{0\} \times \cdots \{1\}(\text{ nth position}) \times \{0\} \times \cdots =\{(x_n)| x_1,x_2\in ([0,1]\cap \Bbb Q),x_n=1,x_i=0, \text{ for }i\neq 1,2,n\}$$ is countable $\forall n$. In a similar fashion $Q_{nm}$ is countable $m>n$ and $Q=\cup_{i<j=1}^{\infty}Q_{ij}$ is countable(being countable union of countable set, it is similar argument to countability to rational number) dense in $S$. Is my line of argument correct?