Proposition 2.31 of Mumford's algebraic geometry I I'm reading Algebraic Geometry I: Complex Projective Varieties by Mumford and have trouble understanding the proof of proposition 2.31.
Let $p_2$ be a projection, and the following proof is only partial.

(2.31) Proposition.
Let $S \subset \mathbb{C}^n \times \mathbb{C}^m$ be a constructible set. Then $p_2(S)\subset \mathbb{C}^m$ is a constructible set. In particular, if $S$ is a subvariety of $\mathbb{C}^{n+m}$ and $\overline{p_2(S)}$ is the Zariski-closure of the image, then $p_2(S)$ contains a Zariski-open set in $\overline{p_2(S)}$.
Proof.
By taking compositions, we are reduced to the case of the projection
$$p_2:\mathbb{C} \times \mathbb{C}^m \longrightarrow \mathbb{C}^m.$$
By induction on the dimension of $\overline{p_2(S)}$, we see easily that it is enough to prove the special case:

if $S \subset \mathbb{C}^{m+1}$ is a subvariety and $S_0 \subset S$ is Zariski-open, then $p_2(S_0)$ contains a Zariski-open subset of $\overline{p_2(S)}$.

Let $T = \overline{p_2(S)}$, then $T$ is variety. Let the affine coordinates rings of $S$ and $T$ be $R_S$ and $R_T$. Then, there is the canonical monomorphism $$\mathbb{C}[X_2, ... X_{m+1}]/I(T)=R_T \longrightarrow R_S = \mathbb{C}[X_1, ..., X_{m+1}]/I(S).$$ Thus $R_S \cong R_T[X_1]/U$ for some ideal $U$. We distinguish two cases: $U = (0)$ hence $S = \mathbb{C}\times T$, and $U \neq (0)$ hence $\dim S = \dim T$. In the first case, ...

How do we prove the proposition from the special case?
Thank you.
 A: Let $S$ be constructible. Then $S$ can be decomposed as a finite disjoint union of sets of the form $S_i=V_i\cap U_i$, where $V_i$ is an irreducible closed subset and $U_i$ is an open subset which intersects $V_i$ nontrivially. As images commute with unions, we have $p_2(S)=\bigcup p_2(S_i)$, and as the finite union of constructible sets is constructible, it suffices to show that $p_2(S_i)$ is constructible for each $i$.
Now assume $S$ is a subvariety and $S_0$ is an open subset. We wish to prove that $p_2(S_0)$ is constructible. Suppose we know that $p_2(S_0)$ contains an open subset $U$ of $\overline{p_2(S_0)}$. Let $V=\overline{p_2(S_0)}\setminus U$ be the closed complement. We have $p_2(S_0)=U\sqcup (p_2(S_0)\cap V)$, so it suffices to show that $p_2(S_0)\cap V$ is constructible.
As $p_2^{-1}(V)\cap S$ is a proper closed subset of $S$, it can be written as a union of varieties $T_i$. We also know that $\dim \overline{p_2(T_i)} < \dim \overline{p_2(S)}$ for each $i$. Therefore for all $i$, we're in the same scenario as we were originally with $T_i$ in the role of $S$ and $S_0\cap T_i$ in the role of $S_0$, but we also know that we strictly decreased the dimension of $\overline{p_2(T_i)}$. So by induction on the dimension of the closure of the projection, we're done. (The base case of the induction is when the dimension of the closure of the projection is zero, aka the projection is a point - that's constructible.)
