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.
First, a small correction: you say something about $X$, $Y$ and $X\times_Y \operatorname{Spec} k(y)$ being of finite type here, but you really should be talking about $X\to\operatorname{Spec} k$, $Y\to\operatorname{Spec} k$, and $X\times_Y \operatorname{Spec} k(y)\to \operatorname{Spec} k(y)$ being of finite type, because being of finite type is a property of morphisms. When one talks about a scheme having a property of a morphism of schemes like this, it is usually assumed that what one means is the canonical morphism to $\operatorname{Spec} \Bbb Z$ has this property. This is problematic for you because no $\Bbb C$-scheme can be of finite type over $\Bbb Z$ for cardinality reasons, for instance. You also make a conclusion about the finite-type-ness of some scheme based on it being a point, but this is inappropriate: $\operatorname{Spec} k[x_1,\cdots]/(x_1,\cdots)^2$ is a single point, but not finite type over $\operatorname{Spec} k$, for example. Basically, don't forget your base!
Let's remember the definition of a finite type morphism: a morphism of schemes $f:X\to Y$ is called finite type if it's quasi-compact and locally of finite type. Quasi-compact means that the inverse image of a quasi-compact set is again quasi-compact, and locally of finite type means that if we have any two open affine schemes $\operatorname{Spec} A\subset X$ and $\operatorname{Spec} R\subset Y$ with $f(\operatorname{Spec} A)\subset \operatorname{Spec} R$, then the induced map on rings $R\to A$ makes $A$ a finite-type $R$-algebra. (We say a ring map $R\to A$ is of finite type if $A$ is isomorphic to a quotient of $R[x_1,\cdots,x_n]$ as an $R$-algebra.)
We'll deal with being locally of finite type first. To be specific:
Lemma (ref). Suppose $X\to Y$ is a morphism of schemes over some base $S$. If $X$ is locally of finite type over $S$, then $X\to Y$ is locally of finite type.
Proof. The condition on rings is equivalent to asking that if $A\to B \to C$ is a sequence of ring maps so that $C$ is finitely generated over $A$, then it's finitely generated over $B$. This is straightforwards: write $C=A[x_1,\cdots,x_n]/J$ and suppose $B$ is generated as an $A$-algebra by some collection of elements $\{y_\alpha\}_{\alpha\in A}$. Let $\overline{y_\alpha}$ denote the image of $y_\alpha$ in $C$. Now I claim that $B[x_1,\cdots,x_n]/(J,y_\alpha-\overline{y_\alpha})\cong C$, where I mean the ideal generated by the images of all elements of $J$ in $B$ and all elements of the form $y_\alpha-\overline{y_\alpha}$ as $\alpha$ ranges over the index set $A$. $\blacksquare$
This previuous lemma is totally general, which is nice! On the other hand, it is not true in general that if $X\to Y$ is a morphism of schemes over a base $S$ and $X\to S$, $Y\to S$ are quasi-compact then one has $X\to Y$ quasi-compact. Examples of this necessarily involve the failure of $Y\to S$ to be quasi-separated, which is probably something you won't see in nature for a while if you're a newer algebraic geometer. (Such an example is necessarily a non-Noetherian scheme, for instance, so if you aren't venturing out of the garden of Noetherian schemes, you're fine.)
In our case where we work over a field, we may conclude the proof as follows. Since $X$ is finite type over a field, it's a noetherian topological space, so every subset of it is quasicompact. This implies that every morphism out of $X$ is quasicompact: the preimage of any set under any morphism coming out of $X$ will be quasicompact, so the definition of a quasicompact morphism is trivially satisfied. Thus, if $X$ and $Y$ are schemes of finite type over a field, then any morphism $X\to Y$ is also of finite type.
Best Answer
Showing that a morphism of schemes is surjective is equivalent to showing that the fibre over every point is nonempty. So you want to consider the pullback of $f^\prime$ along the map $\text{Spec}(k(y)) \rightarrow Y^\prime$ for every $y \in Y^\prime$. Thus you have the following squares:
As both squares are pullback squares by definition, the outer square is also a pullback and thus it suffices to show that the pullback of $f$ along $ \text{Spec}(k(y)) \rightarrow Y$ is still surjective as this immediately implies that $\text{Spec}(k(y)) \times_{Y^\prime} (X \times_Y Y^\prime)$ is nonempty, as desired. Combined with the other reduction step you already mentioned, this gives you the claim of the author.