Now let $q$ be any point of $Z$. Then $q\in X\cap H$ for every $H\in \mathbb{P}^{k-1}$ so $q$ gives a section of $\overline{\Gamma}\longrightarrow\mathbb{P}^{k-1}$. Since $Z$ consists of $d$ distinct points where $d$ is the degree of $X$ we conclude that $q$ is a smooth point of $X$. Finally, since $Z$ is the intersection of all $H\in \mathbb{P}^{k-1}$ this implies that the general intersection $X\cap H$ is smooth at $q$. Summarizing, we have a section of the map which generically lies in the smooth locus of the fibres. Since $\overline{\Gamma}$ is irreducible, this implies that the generic fibre is irreducible, i.e, if $H$ is a generic hyperplane containing $L$, then $H\cap X$ is irreducible.

Finally, the same trick as in the curve case also shows us that for any hyperplane $H$, $H\cap X$ must be non-degenerate. Let $Z=H\cap X$, so that $Z$ is a variety of degree $d$ and dimension $k-1$. Let $M$ be the span of $Z$. If $M\neq H$ then pick any point $p\in X\setminus Z$ and let $H'$ be any hyperplane containing $M$ and $p$. Then $H'\cap X$ can't be all of $X$ (since this would contradict the non-degeneracy of $X$), so $Z'=H'\cap X$ must be a subvariety of dimension $k-1$ (more precisely, all components of $Z'$ have dimension $k-1$) and degree $d$. But $Z$ is therefore a component of $Z'$, and the equality of degrees tells us that $Z'$ can't have any other components so we must have $Z'=Z$. This contradicts the fact that $p\in Z'$ and $p\not\in Z$.

Now let $q$ be any point of $Z$. Then $q\in X\cap H$ for every $H\in \mathbb{P}^{k-1}$ so $q$ gives a section of $\overline{\Gamma}\longrightarrow\mathbb{P}^{k-1}$. Since $Z$ consists of $d$ distinct points where $d$ is the degree of $X$ we conclude that $q$ is a smooth point of $X$. Finally, since $Z$ is the intersection of all $X\cap H$ for $H\in \mathbb{P}^{k-1}$ this implies that the general intersection $X\cap H$ is smooth at $q$. Summarizing, we have a section of the map which generically lies in the smooth locus of the fibres. Since $\overline{\Gamma}$ is irreducible, this implies that the generic fibre is irreducible, i.e, if $H$ is a generic hyperplane containing $L$, then $H\cap X$ is irreducible.

Finally, the same trick as in the curve case also shows us that for any hyperplane $H$, $H\cap X$ must be non-degenerate. Let $Y=H\cap X$, so that $Y$ is a variety of degree $d$ and dimension $k-1$. Let $M$ be the span of $Y$. If $M\neq H$ then pick any point $p\in X\setminus Y$ and let $H'$ be any hyperplane containing $M$ and $p$. Then $H'\cap X$ can't be all of $X$ (since this would contradict the non-degeneracy of $X$), so $Y'=H'\cap X$ must be a subvariety of dimension $k-1$ (more precisely, all components of $Y'$ have dimension $k-1$) and degree $d$. But $Y$ is therefore a component of $Y'$, and the equality of degrees tells us that $Y'$ can't have any other components so we must have $Y'=Y$. This contradicts the fact that $p\in Y'$ and $p\not\in Y$.

If we fix $p$, then the set of possible $H$'s satisfying this condition are simply the hyperplanes $H$ containing the span of $L$ and $p$, and this is parameterized by a $\mathbb{P}^{k-2}$. In other words, the fibres of the projection $\Gamma\longrightarrow X\setminus Z$ are $\mathbb{P}^{k-2}$'s. Since $X\setminus Z$ is irreducible, and since $\mathbb{P}^{k-2}$ is irreducible, this implies that $\Gamma$ is irreducible.

Let $\overline{\Gamma}$ be the Zariski-closure of $\Gamma$ in $\mathbb{P}^{k-1}\times X$. Then $\overline{\Gamma}$ is irreducible since $\Gamma$ is. For a fixed $H\in \mathbb{P}^{k-1}$ the fibre of the projection $\overline{\Gamma}\longrightarrow \mathbb{P}^{k-1}$ over $H$ is simply the intersection $X\cap H$, of dimension $k-1$ (and so connected if $k-1\geqslant 1$). Since $\overline{\Gamma}$ is irreducible, and since the fibres are connected, this implies that the generic fibre is also irreducible, i.e, if $H$ is a generic hyperplane containing $L$, then $H\cap X$ is irreducible.

However, the same trick as in the curve case also shows us that for any hyperplane $H$, $H\cap X$ must be non-degenerate. Let $Z=H\cap X$, so that $Z$ is a variety of degree $d$ and dimension $k-1$. Let $M$ be the span of $Z$. If $M\neq H$ then pick any point $p\in X\setminus Z$ and let $H'$ be any hyperplane containing $M$ and $p$. Then $H'\cap X$ can't be all of $X$ (since this would contradict the non-degeneracy of $X$), so $Z'=H'\cap X$ must be a subvariety of dimension $k-1$ (more precisely, all components of $Z'$ have dimension $k-1$) and degree $d$. But $Z$ is therefore a component of $Z'$, and the equality of degrees tells us that $Z'$ can't have any other components so we must have $Z'=Z$. This contradicts the fact that $p\in Z'$ and $p\not\in Z$.

If we fix $p$, then the set of possible $H$'s satisfying this condition are simply the hyperplanes $H$ containing the span of $L$ and $p$, and this is parameterized by a $\mathbb{P}^{k-2}$. In other words, $\Gamma$ is a $\mathbb{P}^{k-2}$ bundle over $X\setminus Z$. (This fibration is where we use $k\geqslant 2$.) Since $X\setminus Z$ is irreducible this implies that $\Gamma$ is irreducible.

Let $\overline{\Gamma}$ be the Zariski-closure of $\Gamma$ in $\mathbb{P}^{k-1}\times X$. Then $\overline{\Gamma}$ is irreducible since $\Gamma$ is. For a fixed $H\in \mathbb{P}^{k-1}$ the fibre of the projection $\overline{\Gamma}\longrightarrow \mathbb{P}^{k-1}$ over $H$ is simply the intersection $X\cap H$, of dimension $k-1$.

Now let $q$ be any point of $Z$. Then $q\in X\cap H$ for every $H\in \mathbb{P}^{k-1}$ so $q$ gives a section of $\overline{\Gamma}\longrightarrow\mathbb{P}^{k-1}$. Since $Z$ consists of $d$ distinct points where $d$ is the degree of $X$ we conclude that $q$ is a smooth point of $X$. Finally, since $Z$ is the intersection of all $H\in \mathbb{P}^{k-1}$ this implies that the general intersection $X\cap H$ is smooth at $q$. Summarizing, we have a section of the map which generically lies in the smooth locus of the fibres. Since $\overline{\Gamma}$ is irreducible, this implies that the generic fibre is irreducible, i.e, if $H$ is a generic hyperplane containing $L$, then $H\cap X$ is irreducible.

(The intuitive reason for this implication is that, generically over $\mathbb{P}^{k-1}$ the section lets us pick out precisely one irreducible component of the fibre. The union of these components gives us a subset of $\overline{\Gamma}$ which has the same dimension as $\overline{\Gamma}$, and hence whose closure must be all of $\overline{\Gamma}$ by irreducibility. But if there is more than one component in a general fibre, this is a contradiction, thus the general fibre must be irreducible. To make this intuitive construction rigorous requires passing to the normalization of $\overline{\Gamma}$ and then looking at the Stein factorization of the map from the normalization to $\mathbb{P}^{k-1}$. The section gives a generic section of the finite part of the Stein factorization, and that allows one to construct the ``union of the components containing the section''.)

Finally, the same trick as in the curve case also shows us that for any hyperplane $H$, $H\cap X$ must be non-degenerate. Let $Z=H\cap X$, so that $Z$ is a variety of degree $d$ and dimension $k-1$. Let $M$ be the span of $Z$. If $M\neq H$ then pick any point $p\in X\setminus Z$ and let $H'$ be any hyperplane containing $M$ and $p$. Then $H'\cap X$ can't be all of $X$ (since this would contradict the non-degeneracy of $X$), so $Z'=H'\cap X$ must be a subvariety of dimension $k-1$ (more precisely, all components of $Z'$ have dimension $k-1$) and degree $d$. But $Z$ is therefore a component of $Z'$, and the equality of degrees tells us that $Z'$ can't have any other components so we must have $Z'=Z$. This contradicts the fact that $p\in Z'$ and $p\not\in Z$.