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Take the projective line with a doubled point over $\mathbb{C}$, obtained by gluing two copies of ${\mathbb{P}}_{\mathbb{C}}^1$ along the complement of one point. After every base change of a morphism of schemes, a closed subset has closed image from each of the two projective-line charts, so its total image, the union of those two images, is closed. The structure morphism is therefore universally closed. The two doubled points have no disjoint neighborhoods, so the scheme is not separated and hence is not proper.

The coincidence locus of two scheme morphisms is the fibre productThe diagonal morphism ${\Delta }_Y$ is a locally closed immersion, and this property is stable under base change of a morphism of schemes, so $E\hookrightarrow X$ is locally closed. The universal property of a fibre product says that a morphism $T\to X$ factors through $E$ exactly when $f{|}_T=g{|}_T$. Thus $E$ is the largest locally closed subscheme on which they coincide. If $Y$ is separated, ${\Delta }_Y$ is a closed immersion, and its base change $E\hookrightarrow X$ is closed.

A morphism of schemes is proper when it is of finite type, separated, and universally closed.

Each property is local on the target. Finite type is local by its affine definition. The restrictions of the diagonal morphism of $\phi$ over the open sets $U_i$ are closed immersions; since being a closed subset is local on an open cover, the diagonal itself is a closed immersion, so $\phi$ is separated. Finally, after any base change $Y^{\prime }\to Y$, the inverse images $U_i^{\prime }=Y^{\prime }{\times }_Y{U}_i$ cover $Y^{\prime }$. For every closed $C\subseteq X{\times }_Y{Y}^{\prime }$, its image has closed intersection with every $U_i^{\prime }$ because the restricted base-changed morphism is closed. The image is therefore closed in $Y^{\prime }$. Thus $\phi$ is universally closed and hence proper. This proves that properness is local on the target.