If $X$ is an algebraic space such that $X \to S$ is affine (i.e., for every affine scheme $\mathrm{Spec}(A) \to S$, the fiber product $X \times_S \mathrm{Spec}(A)$ is affine), then $X \to S$ is separated. Indeed, affine morphisms are separated.

An algebraic space $X$ equipped with a closed immersion $X \hookrightarrow \mathbb{P}^n_S$ is separated over $S$, since $\mathbb{P}^n_S \to S$ is separated and closed immersions are separated.

Let $U$ be a separated scheme over $S$ and $G$ a finite group acting freely on $U$ over $S$. Then $X = U/G$ is separated over $S$. The diagonal $X \to X \times_S X$ is a closed immersion because $R = G \times U \to U \times_S U$ is a closed immersion (freeness makes this a monomorphism, and finiteness makes it proper, hence closed).

The line with doubled origin $X$ is not separated over $k$. Consider the valuation ring $A = k[[t]]$ with fraction field $K = k((t))$. The map $\mathrm{Spec}(K) \to X$ sending the generic point to $t \in \mathbb{A}^1_k \setminus \{0\}$ admits two extensions to $\mathrm{Spec}(A) \to X$, corresponding to the two origins. This violates the valuative criterion.

Hironaka's example provides a smooth proper threefold $X$ over $\mathbb{C}$ that is separated but not projective. The quotient of $X$ by a free involution gives a proper separated algebraic space that admits no ample line bundle and is thus not a scheme.

Let $E$ be an elliptic curve over $\mathbb{Q}$ and $\sigma$ the negation involution. Away from the 2-torsion points, $\sigma$ acts freely. The quotient of $E \setminus E[2]$ by $\sigma$ is a separated algebraic space over $\mathbb{Q}$, isomorphic to $\mathbb{P}^1 \setminus \{4 \text{ points}\}$.

Every proper scheme over $S$ is a proper algebraic space over $S$. For instance, $\mathbb{P}^n_S \to S$ is proper.

Let $X$ be a proper scheme over $S$ and $G$ a finite group acting freely on $X$ over $S$. Then $X/G$ is a proper algebraic space over $S$. Properness of $X/G$ follows from: (a) it is separated (see above), (b) it is of finite type since $X$ is, and (c) universal closedness is inherited because $X \to X/G$ is surjective and finite.

The quotient of Hironaka's threefold by a free involution is a smooth proper algebraic space of dimension 3 over $\mathbb{C}$ that is not a scheme. It satisfies the valuative criterion because the original threefold is proper, and properness descends along finite surjective maps.

Every proper algebraic space of dimension $\leq 1$ over a field is a projective scheme. This follows from the fact that proper curves have ample line bundles (by Riemann-Roch), so they are projective, and hence schemes. In dimension 1, there is no distinction between proper algebraic spaces and proper schemes.

The coarse moduli space $M_g$ of smooth genus-$g$ curves (for $g \geq 2$) is a proper algebraic space (and in fact a quasi-projective scheme). It is obtained from the proper DM stack $\overline{\mathcal{M}}_g$ via the Keel-Mori theorem.

Over $\mathbb{C}$, there exist proper smooth algebraic spaces that are not projective (and hence not schemes). These correspond to Moishezon manifolds that are not Kähler. The simplest examples occur in dimension 3.

A finite morphism $f: X \to Y$ is proper. Indeed, finite morphisms are affine (hence separated), of finite type, and universally closed (because finite morphisms are closed and this property is stable under base change).

The open immersion $\mathbb{A}^1_k \setminus \{0\} \hookrightarrow \mathbb{A}^1_k$ is separated but not proper (it is not universally closed). The valuative criterion fails: the map $\mathrm{Spec}(k((t))) \to \mathbb{A}^1 \setminus \{0\}$ corresponding to $t^{-1}$ does not extend to $\mathrm{Spec}(k[[t]])$.

For the line with doubled origin $X$, the diagonal $\Delta: X \to X \times X$ is not a closed immersion. The fiber product $X \times X$ has extra points coming from mixed pairs (one origin from each copy), and the diagonal does not capture these. Explicitly, the complement of the diagonal image is not open.

To verify that $X/G \to \mathrm{Spec}(k)$ is proper (for $G$ finite acting freely on proper $X$): (1) Separated: $G \times X \hookrightarrow X \times X$ is a closed immersion since $G$ is finite and the action is free. (2) Finite type: $X$ is of finite type, and $X \to X/G$ is finite surjective, so $X/G$ is of finite type. (3) Universally closed: for any $T$, the map $|X \times_k T| \to |T|$ is closed (properness of $X$), and $|X/G \times_k T| \to |T|$ is the image of a closed map under a finite surjection, hence closed.