Curves which are not universally closed are affine (Hartshorne IV.1.3) I’m working on exercise IV.1.3 in Hartshorne:

Let $X$ be an integral, separated, regular, one-dimensional scheme of finite type over $k$ which is not proper over $k$. Prove $X$ is affine.

In other words, $X$ satisfies all the hypotheses of an abstract curve except that it’s not universally closed over $k$.
Hartshorne’s hint is to embed $X$ in a projective curve $\overline{X}$. I’m assuming one should embed $X$ in projective space and then take the closure. But why should such an embedding exist?
 A: If you don't want to use Nagata compactification, you can use the Riemann–Zariski space in Ch. I, §6 as a compactification. The issue is that you have to mess around a bit with the functor $t\colon \mathfrak{Var}(k) \to \mathfrak{Sch}(k)$. This can probably be solved by allowing the trivial valuation to appear in the definition in Ch. I, §6, which corresponds to the generic point, and so we would not have to leave the world of schemes. For expediency, however, we will just check everything works out correctly when the functor $t$ is applied.
Note that this does not work in higher dimension, since the Riemann–Zariski space (even for a surface) gives you more points than those on the scheme; see Exc. II.4.12. 
Lemma.
Let $X$ be an integral, separated, regular, one-dimensional scheme of finite
  type over an algebraically closed field $k$, of dimension one. Then, $X$ is
  can be identified with an open subset of a projective curve over $k$.
Proof.
  We want to show that $X$ is isomorphic to $t(U)$, where $U$ is an open subset
  of $C_K$, the abstract nonsingular curve of (Ch. I, §6), and $t$ is the
  functor from Prop. II.2.6; we would then be done, since $C_K$ is projective by
  Thm. I.6.9, so $t(C_K)$ is projective by Prop. II.4.10. Since the functor $t$
  is fully faithful, it suffices to show $X(k)$ is isomorphic to some open
  subset $U$ of $C_K$.
By Exercise II.4.5(a), we see that every discrete valuation ring of $K/k$
  has a unique center on $X$, if a center exists. Since the local rings
  of $X$ at the closed points are all DVR's, this implies that
  $X(k)$ is in 1-1 correspondence with a subset of the
  DVR's of $K/k$, namely, a subset $U$ of the points of $C_K$. This subset $U$
  is open by following the proof of Prop. I.6.7.
We have therefore shown that $U$ is an open subset of $C_K$; it remains
  to show that $X(k)$ is isomorphic to $U$.
  It suffices to show that the sheaves of regular functions are the same, since
  the topologies are both the finite complement topology, and they are in
  bijection to each other by the previous paragraph. But this
  follows by the definition of a regular function on $C_K$ on p. 42, since for
  every open $V \subset X(k)$, we have
  $$
    \mathcal{O}_{X(k)}(V) = \bigcap_{P \in V} \mathcal{O}_{X(k),P} = \mathcal{O}_{C_K}(V). \tag*{$\blacksquare$}
  $$
A: There is a remark on page 105, number 4.10.2, which includes mention of a paper by Nagata proving that every variety $X$ can be embedded as an open and dense subset of a complete variety $\overline X$. Using the results of Chapter I.6, mostly Theorem 6.9, you can deduce that $\overline X$ is actually projective. 
