The proof is quite easy. Let $G$ be a simply connected (or adjoint) semisimple group over a field $k$. Over the separable closure $k^{s}$ of $k$, $G$ is a product $G=G_{1}\times\cdots\times G_{n}$ of almost-simple groups $G_{i}$. The Galois group $\Gamma$ of $k^{s}/k$ acts on the set $\{G_{1},\ldots,G_{n}\}$ and the product of the groups in an orbit is stable under $\Gamma$, and hence defined over $k$. In this way, $G$ is a product of quasi-simple groups over $k$. Thus, we may suppose that $G$ itself is quasi-simple. Now $\Gamma$ acts transitively on the set $\{G_{1},\ldots,G_{n}\}$. Let $\Delta$ be the stabilizer of $G_{1}$, and let $K$ be the subfield of $k^{s}$ fixed by $\Delta$. Then $Res_{K/k}(G_{1})$ and $G$ are isomorphic over $K$, by an isomorphism invariant under the action of of $\Delta$, and so they are isomorphic over $k$.

The proof is quite easy. Let $G$ be a simply connected (or adjoint) semisimple group over a field $k$. Over the separable closure $k^{s}$ of $k$, $G$ is a product $G=G_{1}\times\cdots\times G_{n}$ of almost-simple groups $G_{i}$. The Galois group $\Gamma$ of $k^{s}/k$ acts on the set $\{G_{1},\ldots,G_{n}\}$ and the product of the groups in an orbit is stable under $\Gamma$, and hence defined over $k$. In this way, $G$ is a product of quasi-simple groups over $k$. Thus, we may suppose that $G$ itself is quasi-simple. Now $\Gamma$ acts transitively on the set $\{G_{1},\ldots,G_{n}\}$. Let $\Delta$ be the stabilizer of $G_{1}$, and let $K$ be the subfield of $k^{s}$ fixed by $\Delta$. Then $Res_{K/k}(G_{1})$ and $G$ are isomorphic over $k^s$, by an isomorphism invariant under the action of of $\Gamma$, and so they are isomorphic over $k$.