Sorry for resurrecting such an old post, but Section §1.6 of Hartshorne annoys me in the sense that the concept of "abstract nonsingular curve" is not used very much (if at all) in algebraic geometry. We provide an alternative version of that section (specifically, Lemma 1.6.5 through Theorem 1.6.9) that avoids use of said object.
We start with a replacement of Proposition 1.6.8.
Proposition. Let $\phi: X \to Y$ be a rational map from a normal curve $X$ to a projective variety $Y$. Then $\phi$ extends to a (unique) morphism $X \to Y$.
Proof. We may assume that $\phi$ is dominant (replace $Y$ with $\overline{\phi(X)}$), that $Y \subseteq \mathbb{P}^n$, and that $Y$ is not contained in any of the coordinate hyperplanes. Let $U$ be an open subset of $X$ on which $\phi$ is represented by a morphism, and let $P \in X \setminus U$. Since $X$ is normal, $\mathcal{O}_{P,X}$ is a DVR; let $v$ be the associated discrete valuation on $K(X)$. We have that $1 = \phi^*(x_0/x_0), \phi^*(x_1/x_0), \dots, \phi^*(x_n/x_0)$ are nonzero elements of $K(X)$; pick $i$ such that $v(\phi^*(x_i/x_0))$ is minimal. Then, for all $j$,$$v\left(\phi^*\left({{x_j}\over{x_i}}\right)\right) = v\left(\phi^*\left({{x_j}\over{x_0}}\right)\right) - v\left(\phi^*\left({{x_i}\over{x_0}}\right)\right) \ge 0,$$so $\phi^*(x_j/x_i)$ lies in $\mathcal{O}_{P, X}$ for all $j$. Let $V$ be an open neighborhood of $P$ in $X$ such that $\phi^*(x_j/x_i)$ is regular on $V$ for all $j$. Letting $U_i \subseteq \mathbb{P}^n$ denote the open affine $\{x_i \neq 0\}$, we have that the image of the composition$$A(Y \cap U_i) \hookrightarrow K(Y) \overset{\phi^*}{\to} K(X)$$is contained in $\mathcal{O}(V)$, so $\phi^*$ determines a morphism $\phi': V \to Y$ that induces the same map $K(Y) \to K(X)$ as $\phi$. Therefore, $\phi'$ and $\phi$ are the same rational map, and therefore $V$ is contained in the maximal domain of $\phi$. In this way, $\phi$ can be extended to a morphism $X \to Y$. $$\tag*{$\square$}$$
Lemma. Let $W$ be a variety. Then its normalization map $\pi: \widetilde{W} \to W$ is surjective.
Proof. Let $P \in W$. By the lying-over theorem (Atiyah-MacDonald Theorem 5.10 or Eisenbud Proposition 4.15), since $A(\widetilde{W})$ is integral over $A(W)$, it contains a prime ideal whose restriction to $A(W)$ is the maximal (hence prime) ideal $I(P)$. Replacing this prime with a maximal ideal containing it, if necessary we have a maximal ideal $\mathfrak{a}$ for which $\mathfrak{a} \cap A(W) = I(P)$. This maximal ideal corresponds to a point lying over $P$. $$\tag*{$\square$}$$
Theorem. Every curve is birational to a nonsingular projective curve. Moreover, if $C$ is a curve and $\phi: C \to \widetilde{C}$ is a birational map to a nonsingular projective curve $\widetilde{C}$, then $\phi$ is represented by a morphism on the (open) set of nonsingular points of $C$.
Proof. Let $C$ be a curve. We may assume that it is quasi-projective. It is birational to its projective closure, so we may assume that $C$ is projective. Let $\{U_1, \dots, U_m\}$ be a covering of $C$ by open affines. For each $i$, let $\pi_i: V_i \to U_i$ be the normalization (Exercise 1.3.17). This map is birational (since $A(U_i) \subseteq A(V_i) \subseteq K(U_i)$), so there are nonempty open subsets $U_i' \subseteq U_i$ and $V_i' \subseteq V_i$ such that $\pi_i$ induces an isomorphism $V_i' \overset{\cong}{\to} U_i'$.
Each $V_i$ is affine, say $V$ is closed in $\mathbb{A}^{n_i}$ for all $i$. Let $Y_i$ be its projective closure in $P^{n_i}$. We have a rational map$$\theta: C\,\, -\!\! \rightarrow Y_1 \times \dots \times Y_m \subseteq \mathbb{P}^{n_1} \times \dots \times \mathbb{P}^{n_m},$$represented by a morphism on the set $U := \bigcap U_i'$. Let $Y$ be the closure of the graph. Then $Y$ is a projective curve, and we have morphisms $\phi: Y \to C$ and $\psi_i : Y \to Y_i$ for all $i$. As noted earlier, $\phi$ is birational, and is an isomorphism over $U$. In particular, $Y$ is birational to $C$.
In addition, $Y \cap (U \times \prod \pi_i^{-1} (U))$ is the (closed) graph of the morphism$$(\pi_1^{-1}, \dots, \pi_m^{-1}) : U \to \prod Y_i,$$so this nonempty open subset of $Y$ is isomorphic to $U$ and to $\pi_i^{-1}(U)$ for all $i$; hence all $\psi_i$ are also birational. Moreover, we have commutative diagrams
in which the maps on the left induce the maps on the right.
It remains only to show that $Y$ is nonsingular.
Pick $P \in Y$, and choose $i$ such that $\phi(P) \in U_i$. Let $W$ be an open affine neighborhood of $P$ in $\phi^{-1}(U_i)$, and let $\alpha: \widetilde{W} \to W$ be its normalization. By the lemma, there is a point $\tilde{P} \in \widetilde{W}$ such that $\alpha(\widetilde{P}) = P$. We have a commutative diagram
in which $\beta$ exists because $\widetilde{W}$ is normal, $\phi \circ \alpha$ is dominant, and $\pi_i$ is the normalization. However, $\beta$ coincides with $\psi_i \circ \alpha$ on some nonempty open set, since they are both associated to the same map of function fields. Therefore, $\beta = \psi_i \circ \alpha$, so in particular, $\psi_i(P) = \psi_i(\alpha(\tilde{P})) = \beta(\tilde{P})$ lies in $V_i$
Therefore, the local ring $\mathcal{O}_{P, Y}$ dominates $\mathcal{O}_{\psi_i(P), V_i}$. But the latter is normal, hence a DVR (by Theorem 6.2A), hence a valuation ring. By maximality of valuation rings with respect to dominance (Theorem 6.1A), $\mathcal{O}_{P, Y}$ equals $\mathcal{O}_{\psi_i(P), V_i}$; in particular it is regular (by 6.2A again).
This implies the first assertion of the theorem. The second assertion follows immediately from the proposition.$$\tag*{$\square$}$$
Best Answer
You're right, there's some amount of leg work to track down an actual example of a curve here. Let's go through why this result is possible and how to make the exploration as easy as we can.
The way to check whether a nonsingular rational curve $X$ of degree $d$ in $\Bbb P^3$ must be contained in a surface of degree $e$ is twisting the exact sequence $0\to \mathcal{I}_X\to\mathcal{O}_{\Bbb P^3}\to\mathcal{O}_X\to 0$ and then taking global sections: twisting by $e$ gives you $$0\to\mathcal{I}_X(e)\to\mathcal{O}_{\Bbb P^3}(e)\to\mathcal{O}_X(e)\cong\mathcal{O}_{\Bbb P^1}(de)\to 0.$$ The global sections of $\mathcal{O}_{\Bbb P^1}(de)$ are of dimension $de+1$, while the global sections of $\mathcal{O}_{\Bbb P^3}(e)$ are of dimension $\binom{3+e}{3}$. When $d=5$ and $e=3$, we have a map from a vector space of dimension $20$ to a vector space of dimension $16$, so $\Gamma(\Bbb P^3,\mathcal{I}_X(3))\neq 0$ and $X$ lies on a cubic surface; when $d=5$ and $e=3$, we have a map from a vector space of dimension $10$ to a vector space of dimension $11$, so we could have $\Gamma(\Bbb P^3,\mathcal{I}_X(3))=0$ (i.e. $X$ is not on a quadric surface), but we're not locked in to this outcome without doing more computations further on in the long exact sequence of cohomology.
To explain how to set up the guessing process to make it as easy as possible, let's start with what we know: the map $\Bbb P^1\to\Bbb P^3$ is
The fact the morphism is given by $\mathcal{O}(5)$ means it's given by $[a:b:c:d]$ for $a,b,c,d$ quintic polynomials in the coordinates $[s:t]$ on $\Bbb P^1$, and they can't have any common vanishing locus. The easiest way to get that to happen is if $a=s^5$ and $d=t^5$. The fact that this is a closed immersion means that we probably want one of $b$ or $c$ to be $s^4t$ or $st^4$ so that we get that our map is a closed immersion on $D(s)$ or $D(t)$, respectively, and then put a little more work in to showing that our map is a closed immersion on the other patch. This leaves us with three of the entries of our map picked and one to play with:
$$ [s^5: s^4t: ? : t^5]$$
Since we can subtract off any multiples of $s^5$, $s^4t$, or $t^5$ from the $?$ by means of a linear automorphism of $\Bbb P^3$, we might as well assume that $?$ is of the form $ps^3t^2+qs^2t^3+rst^4$. We'll want $r\neq 0$ to get something of valuation $1$ in the local ring at $s=0$ in order to get a closed immersion there, so set $r=1$. Now we need to do two things: find some $p,q$ so that the only quadric vanishing on our curve is $0$, and double check that with those $p,q$ we get a closed immersion on $D(t)\subset \Bbb P^1$. I'll spoiler-tag those calculations so you get a fair shot at doing them yourself. (Personally, I'd say these are potentially a little tedious depending on your appetite for this sort of thing, but they should be doable, and there's some fun in trying to set everything up to give you the best shot at the least work.)
Finding $p,q$:
Checking we get a closed immersion: