Choose a scheme $V'$ and a surjective \'etale morphism $V' \to Y'$. Choose a scheme $U'$ and a surjective \'etale morphism $U' \to X' \times_{Y'} V'$. Set $V = Y \times_{Y'} V'$ and $U = X \times_{X'} U'$. Then for \'etale local properties of morphisms we can reduce to the morphism of thickenings of schemes $(U \subset U') \to (V \subset V')$ and apply More on Morphisms, Lemma \ref{more-morphisms-lemma-thicken-property-morphisms-cartesian}. This proves (2), (3), (4), (5), and (6). \medskip\noindent The properties of morphisms in (1), (7), (8), (9), (10) are stable under base change, hence if $f'$ has property $\mathcal{P}$, then so does $f$. See Spaces, Lemma \ref{spaces-lemma-base-change-immersions}, and Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-base-change-proper}, \ref{spaces-morphisms-lemma-base-change-integral}, and \ref{spaces-morphisms-lemma-base-change-monomorphism}. \medskip\noindent The interesting direction in (1), (7), (8), (9), (10) is to assume that $f$ has the property and deduce that $f'$ has it too. By induction on the order of the thickening we may assume that $Y \subset Y'$ is a first order thickening, see discussion on finite order thickenings above. \medskip\noindent Proof of (1). Choose a scheme $V'$ and a surjective \'etale morphism $V' \to Y'$. Set $V = Y \times_{Y'} V'$, $U' = X' \times_{Y'} V'$ and $U = X \times_Y V$. Then $U \to V$ is a closed immersion, which implies that $U$ is a scheme, which in turn implies that $U'$ is a scheme (Lemma \ref{lemma-thickening-scheme}). Thus we can apply the lemma in the case of schemes (More on Morphisms, Lemma \ref{more-morphisms-lemma-thicken-property-morphisms-cartesian}) to $(U \subset U') \to (V \subset V')$ to conclude. \medskip\noindent Proof of (7). Follows by combining (2) with results of Lemma \ref{lemma-thicken-property-morphisms} and the fact that proper equals quasi-compact $+$ separated $+$ locally of finite type $+$ universally closed. \medskip\noindent Proof of (8). Follows by combining (2) with results of Lemma \ref{lemma-thicken-property-morphisms} and using the fact that finite equals integral $+$ locally of finite type (Morphisms, Lemma \ref{morphisms-lemma-finite-integral}). \medskip\noindent Proof of (9). As $f$ is a monomorphism we have $X = X \times_Y X$. We may apply the results proved so far to the morphism of thickenings $(X \subset X') \to (X \times_Y X \subset X' \times_{Y'} X')$. We conclude $X' \to X' \times_{Y'} X'$ is a closed immersion by (1). In fact, it is a first order thickening as the ideal defining the closed immersion $X' \to X' \times_{Y'} X'$ is contained in the pullback of the ideal $\mathcal{I} \subset \mathcal{O}_{Y'}$ cutting out $Y$ in $Y'$. Indeed, $X = X \times_Y X = (X' \times_{Y'} X') \times_{Y'} Y$ is contained in $X'$. The conormal sheaf of the closed immersion $\Delta : X' \to X' \times_{Y'} X'$ is equal to $\Omega_{X'/Y'}$ (this is the analogue of Morphisms, Lemma \ref{morphisms-lemma-differentials-diagonal} for algebraic spaces and follows either by \'etale localization or by combining Lemmas \ref{lemma-differentials-relative-immersion-section} and \ref{lemma-differential-product}; some details omitted). Thus it suffices to show that $\Omega_{X'/Y'} = 0$ which follows from (5) and the corresponding statement for $X/Y$. \medskip\noindent Proof of (10). If $f : X \to Y$ is an immersion, then it factors as $X \to V \to Y$ where $V \to Y$ is an open subspace and $X \to V$ is a closed immersion, see Morphisms of Spaces, Remark \ref{spaces-morphisms-remark-immersion}. Let $V' \subset Y'$ be the open subspace whose underlying topological space $|V'|$ is the same as $|V| \subset |Y| = |Y'|$. Then $X' \to Y'$ factors through $V'$ and we conclude that $X' \to V'$ is a closed immersion by part (1). This finishes the proof.