Clearly (1) implies (2), (2) implies (3), and (3) implies (4).  It remains to show (4) implies (1). Assume (4). We begin by reducing to $S$ affine.  Being separated is a local on the base (see Schemes, Lemma \ref{schemes-lemma-characterize-separated}). Hence, if we can show that whenever $X \to S$ has (4) that the restriction $X_\alpha \to S_\alpha$ has (4) where $S_\alpha \subset S$ is an (affine) open subset and $X_\alpha := f^{-1}(S_\alpha)$, then we will be done.  The generic points of the irreducible components of $X_\alpha$ will be the generic points of irreducible components of $X$, since $X_\alpha$ is open in $X$.  Therefore, any two distinct dotted arrows in the diagram \begin{equation} \label{equation-valuative-alpha} \xymatrix{ \Spec(K) \ar[r] \ar[d] & X_\alpha \ar[d] \\ \Spec(A) \ar[r] \ar@{-->}[ru] & S_\alpha } \end{equation} would then give two distinct arrows in diagram (\ref{equation-valuative}) via the maps $X_\alpha \to X$ and $S_\alpha \to S$, which is a contradiction.  Thus we have reduced to the case $S$ is affine.  We remark that in the course of this reduction, we prove that if $X \to S$ has (4) then the restriction $U \to V$ has (4) for opens $U \subset X$ and $V \subset S$ with $f(U) \subset V$. \medskip\noindent We next wish to reduce to the case $X \to S$ is finite type.  Assume that we know (4) implies (1) when $X$ is finite type. Since $S$ is Noetherian and $X$ is locally of finite type over $S$ we see $X$ is locally Noetherian as well (see Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}). Thus, $X \to S$ is quasi-separated (see Properties, Lemma \ref{properties-lemma-locally-Noetherian-quasi-separated}), and therefore we may apply the valuative criterion to check whether $X$ is separated (see Schemes, Lemma \ref{schemes-lemma-valuative-criterion-separatedness}). Let $X = \bigcup_\alpha X_\alpha$ be an affine open cover of $X$. Given any two dotted arrows, in a diagram (\ref{equation-valuative}), the image of the closed points of $\Spec A$ will fall in two sets $X_\alpha$ and $X_\beta$.  Since $X_\alpha \cup X_\beta$ is open, for topological reasons it must contain the image of $\Spec(A)$ under both maps. Therefore, the two dotted arrows factor through $X_\alpha \cup X_\beta \to X$, which is a scheme of finite type over $S$. Since $X_\alpha \cup X_\beta$ is an open subset of $X$, by our previous remark, $X_\alpha \cup X_\beta$ satisfies (4), so by assumption, is separated.  This implies the two given dotted arrows are the same. Therefore, we have reduced to $X \to S$ is finite type. \medskip\noindent Assume $X \to S$ of finite type and assume (4). Since $X \to S$ is finite type, and $S$ is an affine Noetherian scheme, $X$ is also Noetherian (see Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}). Therefore, $X \to X \times_S X$ will be a quasi-compact immersion of Noetherian schemes.  We proceed by contradiction.  Assume that $X \to X \times_S X$ is not closed.  Then, there is some $y \in X \times_S X$ in the closure of the image that is not in the image. As $X$ is Noetherian it has finitely many irreducible components. Therefore, $y$ is in the closure of the image of one of the irreducible components $X_0 \subset X$.  Give $X_0$ the reduced induced structure.  The composition $X_0 \to X \to X \times_S X$ factors through the closed subscheme $X_0 \times_S X_0 \subset X \times_S X$. Denote the closure of $\Delta(X_0)$ in $X_0 \times_S X_0$ by $\bar X_0$ (again as a reduced closed subscheme). Thus $y \in \bar X_0$. Since $X_0 \to X_0 \times_S X_0$ is an immersion, the image of $X_0$ will be open in $\bar X_0$. Hence $X_0$ and $\bar X_0$ are birational. Since $\bar{X}_0$ is a closed subscheme of a Noetherian scheme, it is Noetherian. Thus, the local ring $\mathcal O_{{\bar X_0, y}}$ is a local Noetherian domain with fraction field $K$ equal to the function field of $X_0$.  By the Krull-Akizuki theorem (see Algebra, Lemma \ref{algebra-lemma-exists-dvr}), there exists a discrete valuation ring $A$ dominating $\mathcal O_{{\bar X_0, y}}$ with fraction field $K$.  This allows to construct a diagram: \begin{equation} \label{equation-valuative-generic} \xymatrix{ \Spec(K) \ar[r] \ar[d] & X_0 \ar[d]^{\Delta} \\ \Spec(A) \ar[r] \ar@{-->}[ur]& X_0 \times_S X_0 \\ } \end{equation} which sends $\Spec K$ to the generic point of $\Delta(X_0)$ and the closed point of $A$ to $y \in X_0 \times_S X_0$ (use the material in Schemes, Section \ref{schemes-section-points} to construct the arrows). There cannot even exist a set theoretic dotted arrow, since $y$ is not in the image of $\Delta$ by our choice of $y$.  By categorical means, the existence of the dotted arrow in the above diagram is equivalent to the uniqueness of the dotted arrow in the following diagram: \begin{equation} \label{equation-valuative-nonexistent} \xymatrix{ \Spec(K) \ar[r] \ar[d] & X_0 \ar[d]\\ \Spec(A) \ar[r] \ar@{-->}[ur] & S \\ } \end{equation} Therefore, we have non-uniqueness in this latter diagram by the nonexistence in the first.  Therefore, $X_0$ does not satisfy uniqueness for discrete valuation rings, and since $X_0$ is an irreducible component of $X$, we have that $X \to S$ does not satisfy (4).  Therefore, we have shown (4) implies (1).