This question concerns a Marvin Greenberg's paper on Rational Points in Henselian Discerete Valuation Rings (springer link) and part of the main's theorem proof on page 62.
Let $(R,m,k)$ be a Henselian DVR and $Y$ be an affine, non-separable scheme over $R$ defined by the polynomial system $I$ ($Y= \operatorname{Spec} R[X_1,...X_n]/I \cdot R[X_1,...X_n]))$. Let $K'$ be a purely inseparable finite extension of field of fractions $K:= Q(R)$ such that the scheme $Y_{K'}$ is not reduced, then a fortiori $Y_{K'}$ is not reduced. (By [EGA IV; 4.6-3] such inseparable $K'$ always exists).
It uses the Fact 2 that There is a functor $\mathcal{F}$ from the category of affine schemes of finite type over $R'$ to affine schemes of finite type over $R$ such that $\mathcal{F}$ is right adjoint to the change of base functor from $R$ to $R'$. Thus we have an isomorphism of bifunctors
$$ \operatorname{Mor}_R(Y, \mathcal{F}Z) \to \operatorname{Mor}_{R'}(Y_{R'}, Z) $$
for $Y/R, Z/R'$. Moreover, $\mathcal{F}$ preserves closed immersions. [Technique de descent, II, Seminaire Bourbaki; p. 195-13] Now we can consider the part of the original proof:
Consider the affine scheme $\mathcal{F}Y_{R'}$ over $R$. There is a canonical $R$-morphism $\theta: Y \to \mathcal{F}Y_{R'}$ which corresponds by adjointness to the identity morphism of $Y_{R'}$. By Fact 2, $\mathcal{F}((Y_{R'})_{red})$ is a closed subscheme of $\mathcal{F}Y_{R'}$; let $W:= \theta^{-1}(\mathcal{F}((Y_{R'})_{red}))$ be its pre-image under $\theta$.
The claim is that then $W$ is a proper closed subcheme of $Y$, otherwise the identity morphism of $Y_{R'}$, would factor through $(Y_{R'})_{red}$, i.e., $Y_{R'}$ would be reduced, contradicting the choice of $R'$.
The last argument is hard to understand. I'm going to assure that $W \neq Y$ and assume for now that $W= Y$ and pursue to get a contradiction.
The composition $W = \theta^{-1}(\mathcal{F}((Y_{R'})_{red})) \to \mathcal{F}((Y_{R'})_{red}) \to \mathcal{F}((Y_{R'})$ coinsides after identification $W= Y$ with $\theta: Y \to \mathcal{F}Y_{R'}$ and since the isomorphism of bifunctors in Fact 2 is functorial I get a factorization of identity morphism of $Y_{R'}$ by $Y_{R'} \to (Y_{R'})_{red} \to Y_{R'}$ where the right arrow is the canonical closed embedding.
Why this implies that $Y_{R'} = (Y_{R'})_{red} $, i.e., that $Y_{R'}$ is reduced? Don't we need also to assure that the identity of $ (Y_{R'})_{red}$ factorizes with same morphisms as $ (Y_{R'})_{red} \to (Y_{R'}) \to (Y_{R'})_{red}$ to complete the proof?