I have some question with my answer and I`m interested in if my answer is correct.

Suppose (X,T) is any completely normal topological space. Lets 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. Than (Y,F) is also completely normal.

Proof: Suppose A and B is seperable sets in (Y,F), so cl(A)∩B=A∩cl(B)=∅, whete cl(*) is closure operator. A and B is seperable sets in Y iff A1≡{x∈X:(x,0)∈A}∪{x∈X:(x,1)∈A} and B1≡{x∈X:(x,0)∈B}∪{x∈X:(x,1)∈B} is seperable sets in X. So, There exist U and V disjoint open sets in X such, that A1⊂U and B1⊂V. Than U×{0,1} and V×{0,1} is disjoint open sets in Y such, that A⊂U×{0,1} and V⊂V×{0,1}. So, (Y,F) is completely normal.

Is my proof correct? Thank you!