A subspace of a T₀ space is T₀.

Proof: Let X be T₀ and let A be a subspace of X. Let x, y &isin A. Then x, y &isin X. Hence there exists a set U which is open in X such that either [x &isin U and y ¬in U] or [y &isin U and x ¬in U]. U &cap A is open in A and either [x &isin U &cap A and y ¬in U &cap A] or [y &isin U &cap A and x ¬in U &cap A]. Thus A is T₀.