Lemma 2

If a metrizable space $X$ is such that there exists a countable partition $(A_n)$ of $X$ formed of Lusin subspaces, then $X$ is a Lusin space.

For each $n$, let $P_n$ be a zero-dimensional Polish space and let $f_n$ be a continuous bijection of $P_n$ onto $A_n$. If $P$ is the topological sum of the $P_n$, then $P$ is Polish (no. 1, Proposition 1) and zero-dimensional, and the mapping $f : P \to X$ which agrees with $f_n$ on $P_n$ for each $n$ is a continuous bijection of $P$ onto $X$; hence the result.

Let us now show that the interval $I = [0, 1]$ of $\mathbf{R}$ is a Lusin space. Let $J$ be the subspace of $I$ consisting of all irrational numbers; then $J$ is Polish (no. 1, Corollary to Proposition 3); also J is zero-dimensional, for if x is any point of J, the traces on J of intervals of R of the form ]r, s[, where r and s are rational and r < x < s, form a fundamental system of open and closed neighbourhoods of x in J (because the traces on J of ]r, s[ and [r, s] are the same). Hence J is a Lusin subspace of I. Now J and the subspaces of I which consist each of a single rational point form a countable partition of I, and therefore I is a Lusin space by Lemma 2.

Let P be an arbitrary Polish space. By Corollary 1 to Theorem 1 (no. 1), P is homeomorphic to a subspace of the cube IN which is a countable intersection of open sets in IN. Since I is a Lusin space, the remarks at the beginning of the proof show that P is a Lusin space, and the proof of Proposition 12 is therefore complete.