You are correct, and something similar holds more generally for Weil restriction for a quasi-projective $X' \rightarrow S'$ through a map $f:S' \rightarrow S$ that is finite locally free of constant rank $d$. (So in the field case your separability hypothesis is unnecessary.)

In fact this is related to an apparent technical gap in the discussion in that book: they don't address why the Weil restriction is again quasi-compact (and hence of finite type) over the base, and its proof seems to require your observation (generalized to the relative base case).

To be precise, the construction in that book for affine $S$ and $S'$ (which is the essential case) is initially done using *all* open affines in $X'$, so a-priori it is a just locally of finite type over $S$. Note that existence as a locally finite type $S$-scheme is sufficient for knowing that ${\rm{R}}_{S'/S}(U') \rightarrow {\rm{R}}_{S'/S}(X')$ is an open immersion for any open subscheme $U'$ of $X'$ because open immersions are the same thing as etale monomorphisms (without any quasi-compactness hypotheses on morphisms!).

Since $X' \rightarrow S'$ is quasi-projective over an affine $S'$, any finite subset of $X'$ lies in an affine open subset. Thus, there does exist a (perhaps infinite) collection $\{U_i\}$ of affine opens such that any ordered $d$-tuple in $X'$ (*allowing repetitions*, for a technical reason to be seen later!) lies in some $U_i$. Now there are two things to be done: (1) prove that for any such $\{U_i\}$ the collection of open subschemes ${\rm{R}}_{S'/S}(U_i)$ covers ${\rm{R}}_{S'/S}(X')$ (a generalization of your guess), (2) show that there is such a collection $\{U_i\}$ that is *finite*.

Assertion (2) is point-set topology without Hausdorffness, as follows. Pick some $\{U_i\}$ as above but perhaps infinite. The *topological* product ${X'}^d$ of all ordered $d$-tuples (allowing repetitions!) is quasi-compact since $X'$ is quasi-compact, and the *open* subsets $(U_i)^d$ do cover it (by the quasi-projectivity, as indicated above), so there is a finite subcover, say $(U_{i_1})^d, \dots, (U_{i_r})^d$ for some $i_1,\dots, i_r$. Thus, $\{U_{i_1},\dots,U_{i_r}\}$ answers (2) affirmatively.

It remains to prove (1), which in effect is the problem you're asking about and for which you have already identified the argument. To be pedantic, here is your argument. A point of a scheme is the image of a morphism from a field, so a point of ${\rm{R}}_{S'/S}(X')$ corresponds to the image of a map $x:{\rm{Spec}}(F) \rightarrow {\rm{R}}_{S'/S}(X')$ for a field $F$. Composing $x$ with the structure map to $S$ defines an $F$-valued point $s: {\rm{Spec}}(F) \rightarrow S$, and by the functorial meaning of Weil restriction we see that $x$ viewed as an $S$-morphism (using $s$) corresponds to an $S'$-morphism $x':S'_s \rightarrow X'$. But $S'_s$ is an $F$-scheme of rank $d$, so it consists of at most $d$ physical points, and hence lands inside one of the open subschemas $U_i$ (because we allow our ordered $d$-tuples to contain repetitions). Now we can run the calculation in reverse with that $U_i$ in the role of $X'$ to deduce that $x$ factors through ${\rm{R}}_{S'/S}(U_i)$.