Nothing is "better-suited to using the classical language"; if you cannot express things clearly via schemes then think harder about it until you can. Also, any connected reductive group over a field has a *unique* quasi-split inner form. (See Proposition 7.2.12 in the article *Reductive Group Schemes* in the Proceedings of the 2011 Luminy Summer School on SGA3 for a proof of the same result more generally over any semi-local base scheme.)

If $k_s/k$ is a separable closure and $H$ is the $k$-split form of $G$, with a split maximal $k$-torus $S$ and Borel $k$-subgroup $B\supset S$, there is a ${\rm{Gal}}(k_s/k)$-equivariant exact sequence of groups $$1 \rightarrow H^{\rm{ad}}(k_s)\to {\rm{Aut}}_{k_s}(H_{k_s})\stackrel{\pi}{\to} {\rm{Aut}}(R,\Delta)\to 1$$ where $(R,\Delta)$ is the based root datum for $(H,B,S)$. A pinning $\{u_{\alpha}\}$ identifies ${\rm{Aut}}(R,\Delta)$ with ${\rm{Aut}}_k(H,S,B,\{u_{\alpha}\})$ and thereby defines a "forgetful" homomorphic section
$$\sigma:{\rm{Aut}}(R,\Delta) \rightarrow {\rm{Aut}}_k(H) \subset
{\rm{Aut}}_{k_s}(H_{k_s})$$
to $\pi$.

Consider the class $[G] \in {\rm{H}}^1(k_s/k, {\rm{Aut}}_{k_s}(H_s))$ corresponding to $G$. The formalism of non-abelian Galois cohomology as in section 5 of Chapter I of Serre's book *Galois Cohomology* shows that the set of isomorphism classes of inner forms of $G$ (i.e., the image of ${\rm{H}}^1(k_s/k, G^{\rm{ad}}(k_s))$ in ${\rm{H}}^1(k_s/k, {\rm{Aut}}_{k_s}(G_{k_s})$) is identified with the ${\rm{H}}^1(\pi)$-fiber through $[G]$.

In particular, ${\rm{H}}^1(\sigma)({\rm{H}}^1(\pi)([G]))$ is a class in the same fiber as $[G]$, so for the existence of a quasi-split inner form of $G$ it suffices to show that all classes in the image of ${\rm{H}}^1(\sigma)$ are quasi-split. But $\sigma$ is defined via the identification of ${\rm{Aut}}(R,\Delta)$
with ${\rm{Aut}}_k(H,S,B, \{u_{\alpha}\})$ and so factors through the subgroup ${\rm{Aut}}_{k_s}(H_{k_s},B_{k_s}) \subset {\rm{Aut}}_{k_s}(H_{k_s})$. Thus, any class in the image of ${\rm{H}}^1(\sigma)$ corresponds to the isomorphism class of a $k$-group $H'$ obtained through ${\rm{Gal}}(k_s/k)$-twisting of $H$ preserving its Borel $k$-subgroup $B$, so by design $H'$ admits a Borel $k$-subgroup $B'$ and thus $H'$ is quasi-split over $k$.

The remaining claim (not posed in the question, but mentioned at the start of this answer and very important in practice) is that the quasi-split inner form is unique. That is, if $G_1$ and $G_2$ are quasi-split inner forms of $G$ then they are $k$-isomorphic. More specifically, if $G_1$ is the quasi-split inner form made via the above construction and $G_2$ is any quasi-split inner form then $G_2 \simeq G_1$. This lies deeper in the sense that it rests on a more closer study of the preceding construction.

First, it is an instructive exercise to prove in a clean way that $G_2$ is necessarily an inner form of $G_1$, so we can rename $G_1$ as $G$ to reduce to showing that if $G$ admits a Borel $k$-subgroup $B$ *with $(G,B)$ built from the split $k$-form via the above procedure* resting on $\sigma$ then any quasi-split *inner form* $G'$ of $G$ is $k$-isomorphic to $G$. For a given choice of $k_s$-isomorphism $f:G'_{k_s} \simeq G_{k_s}$ corresponding to *inner-twisting* cocycles realizing $G'$ from $G$ via Galois descent, if we postcompose with a $G(k_s)$-conjugation we get a cohomologous 1-cocycle that has the same "inner-twisting" property. Thus, it is harmless to arrange that $f(B'_{k_s}) = B_{k_s}$, so then the inner-twisting is valued in the $B_{k_s}$-stabilizer subgroup of $G^{\rm{ad}}(k_s)$. But this stabilizer is exactly $B^{\rm{ad}}(k_s)$ for $B^{\rm{ad}} = B/Z_G$ so it suffices to prove
$${\rm{H}}^1(k,B^{\rm{ad}}) = 1.$$
In this way we can replace $G$ with $G^{\rm{ad}}$ without ruining any of our running hypotheses on $G$ but gaining that we may now assume $G$ is of adjoint type.

The $k$-unipotent radical of any parabolic $k$-subgroup of a connected reductive group is always $k$-split and so has vanishing ${\rm{H}}^1$. Thus, since $B = T \ltimes U$ for a maximal $k$-torus $T$ of $B$ and
$U = \mathscr{R}_{u,k}(B)$, so $B/U \simeq T$, it suffices to show ${\rm{H}}^1(k, T) = 1$ when $G$ is quasi-split *of adjoint type* and made from its split form via the procedure resting on $\sigma$. For this we finally have to make a serious observation that uses that $G$ is of adjoint type and is constructed from its split form via $\sigma$: the $k$-torus $T$ is "induced" (i.e., $T \simeq {\rm{R}}_{k'/k}({\rm{GL}}_1)$ for a finite etale $k$-algebra $k'$), from which it is immediate via Hilbert 90 that the desired ${\rm{H}}^1$-vanishing holds.

Why is $T$ induced? By inspection of how $\sigma$ is made, $T$ is build from Galois-twisting of a split "adjoint torus" $S = {\rm{GL}}_1^{\Delta}$ through Galois action on ${\rm{X}}(S_{k_s}) = \mathbf{Z}^{\Delta}$ via a permutation action on $\Delta$. The Galois-orbits on $\Delta$ then make explicit that the associated $k$-form $T$ of $S$ is an induced torus. (Explicitly, if we pick a point in each Galois-orbit on $\Delta$ then the stabilizer in ${\rm{Gal}}(k_s/k)$ of each such base point is an open subgroup corresponding to a finite separable extension $k'_i/k$, and one shows $T \simeq \prod_i {\rm{R}}_{k'_i/k}({\rm{GL}}_1)$.)