Let $c_1, \ldots, c_n \in k$ be as in the statement of the lemma. Write $\xi = \sum g_ix_i$ and $\delta = \sum c_ig_i$. For $a \in k$ consider the automorphism $\sigma_a$ of $K(x_1, \ldots, x_n)$ given by the identity on $K$ and the rules $$ \sigma_a(x_i) = x_i + a c_i $$ Observe that $\sigma_a(\xi) = \xi + a \delta$ and $\sigma_a(\delta) = \delta$. Consider the tower of fields $$ K_0 = k(x_1, \ldots, x_n) \subset K_1 = K_0(\xi) \subset K_2 = K_0(\xi, \delta) \subset K(x_1, \ldots, x_n) = \Omega $$ Observe that $\sigma_a(K_0) = K_0$ and $\sigma_a(K_2) = K_2$. Let $\theta \in \Omega$ be separable algebraic over $K_1$. We have to show $\theta \in K_1$, see Algebra, Lemma \ref{algebra-lemma-geometrically-irreducible-separable-elements}. \medskip\noindent Denote $K'_2$ the separable algebraic closure of $K_2$ in $\Omega$. Since $K'_2/K_2$ is finite (Algebra, Lemma \ref{algebra-lemma-make-geometrically-irreducible}) and separable there are only a finite number of fields in between $K'_2$ and $K_2$ (Fields, Lemma \ref{fields-lemma-primitive-element}). If $k$ is infinite\footnote{We will deal with the finite field case in the last paragraph of the proof.}, then we can find distinct elements $a_1, a_2$ of $k$ such that $$ K_2(\sigma_{a_1}(\theta)) = K_2(\sigma_{a_2}(\theta)) $$ as subfields of $\Omega$. Write $\theta_i = \sigma_{a_i}(\theta)$ and $\xi_i = \sigma_{a_i}(\xi) = \xi + a_i \delta$. Observe that $$ K_2 = K_0(\xi_1, \xi_2) $$ as we have $\xi_i = \xi + a_i \delta$, $\xi = (a_2 \xi_1 - a_1 \xi_2)/(a_2 - a_1)$, and $\delta = (\xi_1 - \xi_2)/(a_1 - a_2)$. Since $K_2/K_0$ is purely transcendental of degree $2$ we conclude that $\xi_1$ and $\xi_2$ are algebraically indepedent over $K_0$. Since $\theta_1$ is algebraic over $K_0(\xi_1)$ we conclude that $\xi_2$ is transcendental over $K_0(\xi_1, \theta_1)$. \medskip\noindent By assumption $K/k$ is geometrically irreducible. This implies that $K(x_1, \ldots, x_n)/K_0$ is geometrically irreducible (Algebra, Lemma \ref{algebra-lemma-geometrically-irreducible-base-change-transcendental}). This in turn implies that $K_0(\xi_1, \theta_1)/K_0$ is geometrically irreducible as a subextension (Algebra, Lemma \ref{algebra-lemma-subalgebra-geometrically-irreducible}). Since $\xi_2$ is transcendental over $K_0(\xi_1, \theta_1)$ we conclude that $K_0(\xi_1, \xi_2, \theta_1)/K_0(\xi_2)$ is geometrically irreducible (Algebra, Lemma \ref{algebra-lemma-geometrically-irreducible-add-transcendental}). By our choice of $a_1, a_2$ above we have $$ K_0(\xi_1, \xi_2, \theta_1) = K_2(\sigma_{a_1}(\theta)) = K_2(\sigma_{a_2}(\theta)) = K_0(\xi_1, \xi_2, \theta_2) $$ Since $\theta_2$ is separably algebraic over $K_0(\xi_2)$ we conclude by Algebra, Lemma \ref{algebra-lemma-geometrically-irreducible-separable-elements} again that $\theta_2 \in K_0(\xi_2)$. Taking $\sigma_{a_2}^{-1}$ of this relation givens $\theta \in K_0(\xi) = K_1$ as desired. \medskip\noindent This finishes the proof in case $k$ is infinite. If $k$ is finite, then we can choose a variable $t$ and consider the extension $K(t)/k(t)$ which is geometrically irreducible by Algebra, Lemma \ref{algebra-lemma-geometrically-irreducible-base-change-transcendental}. Since it is still be true that $x_1, \ldots, x_n, \sum g_ix_i, \sum c_ig_i$ in $K(t, x_1, \ldots, x_n)$ are algebraically independent over $k(t)$ we conclude that $K(t, x_1, \ldots, x_n)$ is geometrically irreducible over $k(t, x_1, \ldots, x_n, \sum g_ix_i)$ by the argument already given. Then using Algebra, Lemma \ref{algebra-lemma-geometrically-irreducible-base-change-transcendental} once more finishes the job.