The line with doubled origin, and dropping each hypothesis

(Not separated)

Glue two copies of \(\AA^1_k\) along \(\AA^1 \sm \ts{0}\) by the identity. The result \(X\) has two origins, and \(X \to \Spec k\) is not separated: the two inclusions \(\AA^1 \to X\) agree on the dense open \(\AA^1 \sm \ts{0}\) and differ at \(0\), so the agreement locus is not closed.

By the proposition this is the failure of separatedness, not merely an illustration of it.

(Separated but not reduced)

Take \(Z = \Spec k[\varepsilon]/(\varepsilon^2)\) and \(X = \AA^1 = \Spec k[x]\), with \(g\) and \(h\) given by \(x \mapsto 0\) and \(x \mapsto \varepsilon\). Both are morphisms to a separated \(X\), and they agree on the unique point of \(Z\) — indeed on every reduced subscheme of \(Z\) — but \(g \neq h\). The dense open here is not proper, and it need not be: on a nonreduced base, agreeing topologically is not agreeing.