The definition of open subscheme in Hartshorne is so bad that the first sentence he writes after it is false!
That sentence is: Note that every open subset of a scheme carries a unique structure of open subscheme (Ex.2.2).
This is false: indeed, replacing $\mathcal O_{X|U}$ by an isomorphic sheaf you will get another scheme (albeit isomorphic, of course) and the claimed unicity does not hold..
And you really get other schemes because a scheme is a pair formed by a topological space and a sheaf of rings, not an isomorphism class of sheaves of rings.
So that the collection of subscheme structures on an open subset $U$ of a scheme $X$ would not even be a set with Hartshorne's definition ...
Of course in EGA, Görz-Wedhorn, Qing Liu,... you will find the reasonable definition that an open subscheme of $X$ is an open subset $U\subset X$ endowed with the restricted sheaf $\mathcal O_{X|U}$ , and not some arbitrary sheaf isomorphic to it. There is of course a canonical morphism of schemes $j:U\to X$
Edit: Open immersions Once you have the correct notion of open subscheme, you have no choice for defining an open immersion $Y\to X$. The natural idea is that you want it to be an isomorphism $g:Y \stackrel {\sim}{\to} U $ where $U\subset X$ is an open subscheme. However this has the wrong target, so you just compose it with the canonical morphism $j:U\to X$ mentioned above and you get the required immersion $f=j\circ g:Y\to X$, just as Wikipedia and most other references say.
And, quite satisfactorily, $j:U\to X$ itself is then an open immersion ( take $g=id_U$).
A criterion for being an open immersion: In practice to prove that a morphism $f:Y\to X$ is an open immersion, it suffices to check that:
a) $f$ induces a homeomorphism of $Y$ onto an open subset $U$ of $X$.
b) For all $y\in Y$ the local morphism $f^\ast_y: \mathcal O_{X,f(y)} \to \mathcal O_{Y,y}$ is an isomorphism of local rings.
NB Needless to say, I have an immense admiration for Hartshorne: I'm criticizing a definition in a book, and certainly not that great, friendly mathematician!
As stated in your excerpt: $i_U : U \times_{S'} U \to X \times_S X$ is an open embedding.
Choosing such $U_x$ for varying $x \in X$ results in an open cover $X = \bigcup_x U_x$.
As each $U \times_{S'} U$ is open in $X\times_S X$, so is $V := \bigcup_{x}U_x\times_{S'_x}U_x$.
Note however, that it is not clear (and in general not true!), that $X\times_S X = \bigcup_x U_x \times_{S'_x} U_x$.
Hence we got an open embedding $i:V \to X$.
Now as $\Delta_X |_U = i_U \circ \Delta_U$ factorizes over $i_U$, it actually takes values in $U \times_{S'} U$.
Hence $\Delta_X$ factorizes over the open embedding $i: V \to X$, i.e. $\Delta_X = i \circ \Delta_X|^V$, where $\Delta_X|^V : X \to V$ is the "corestriction" to $V$.
It therefore suffices to show that $\Delta_X|^V$ is a closed embedding.
Note that being a closed embedding is a property that is local on the target:
$f : X \to Y$ is a closed embedding iff. $f| : f^{-1} V \to V$ is a closed embedding for any open $V$ in an open cover $Y = \bigcup_V V$.
In our case we have the open cover $V = \bigcup_x U_x \times_{S'_x} U_x$ and we know that $\Delta_X|^V$ restricts exactly to $\Delta_U : U \to U\times_{S'} U$, which is shown to be a closed embedding.
Best Answer
No. Let $x\in\Bbb A^1_k$ be a closed point, and consider $\{x\}\to U\to \Bbb A^1_k$ as $U$ varies among the open subsets of $\Bbb A^1_k$ containing $x$.
Yes. Suppose $W\subset X$ is a locally closed subset. Write $W=U\cap V$ with $U$ open and $V$ closed. Equip $U$ with the restriction of the structure sheaf on $X$ and $V$ with the reduced induced subscheme structure. Then $W=U\times_X V$ is a reduced subscheme structure on $W$.
Hartshorne's definition of a locally closed immersion is famously not quite correct in general. To be more specific, Hartshorne's locally closed immersion is the opposite order from what you've written - an open immersion in to a closed subscheme. One can interchange the order of these when either the source is reduced or the composite morphism is quasi-compact. The latter condition is satisfied when the source is (locally) noetherian, and given Hartshorne's general perspective on noetherian hypotheses, I think the best way to proceed is to assume he's working in the locally noetherian situation. This gives that a scheme $X$ quasi-projective over $S$ via $f:X\to S$ can be written as a locally closed subscheme of $\Bbb P^n_S$ for some $n$. Generally one interprets this as the second condition you've written - there exists a map, but it's not generally packaged as part of the data of $f$.