If a topological space (X,T) is completely normal, and if we double the point of X, is the resulting space also completely normal? [closed]

Question

I have a question on my proof of the following lemma, and I'd like to know if my answer is correct.

Lemma. Suppose $(X,T)$ is any completely normal topological space. Let's double the points of $X$, more precise, consider space $(Y,F)$, where $Y$ is the cartesian product of $X$ and $\{0,1\}$ (with trivial topology) and $F$ is product topology. Then $(Y,F)$ is also completely normal.

Proof: Suppose $A$ and $B$ are separated sets in $(Y,F)$, so $$\operatorname{clo}(A)\cap B=A\cap \operatorname{clo}(B)=\emptyset,$$ where $\operatorname{clo}(\cdot)$ is the closure operator. $A$ and $B$ are separable sets in $Y$ iff $$ A_1\equiv\{x\in X:(x,0)\in A\}\cup\{x\in X:(x,1)\in A\} $$ and $$ B_1\equiv \{x\in X:(x,0)\in B\}\cup\{x\in X:(x,1)\in B\} $$ are separated sets in $X$. So, there exist $U$ and $V$ disjoint open sets in $X$ such, that $A_1\subset U$ and $B_1\subset V$. Then $U\times\{0,1\}$ and $V\times\{0,1\}$ are disjoint open sets in $Y$ such that $A\subset U\times\{0,1\}$ and $V\subset V\times\{0,1\}$. So $(Y,F)$ is completely normal.

Is my proof correct? Thank you!

Answer 1

Apart from the use of the word 'separable' it is correct; the term normally used is 'separated', and separable has the well-established meaning of 'having a countable dense subset'.