By induction on the number of irreducible components of $X$. If $X$ is irreducible, then the lemma holds by Lemma \ref{lemma-degree-more-than-2g} applied to $X$ viewed as a scheme over the field $k' = H^0(X, \mathcal{O}_X)$. Assume $X$ is not irreducible. Before we continue, if $k$ is finite, then we replace $k$ by a purely transcendental extension $K$. This is allowed by Varieties, Lemmas \ref{varieties-lemma-globally-generated-base-change}, \ref{varieties-lemma-degree-base-change}, \ref{varieties-lemma-geometrically-reduced-any-base-change}, and \ref{varieties-lemma-bijection-irreducible-components}, Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}, Lemma \ref{lemma-sanity-check-duality} and the elementary fact that $K$ is geometrically integral over $k$. \medskip\noindent Assume that $\mathcal{L}$ is not globally generated to get a contradiction. Then we may choose a coherent ideal sheaf $\mathcal{I} \subset \mathcal{O}_X$ such that $H^0(X, \mathcal{I}\mathcal{L}) = H^0(X, \mathcal{L})$ and such that $\mathcal{O}_X/\mathcal{I}$ is nonzero with support of dimension $0$. For example, take $\mathcal{I}$ the ideal sheaf of any closed point in the common vanishing locus of the global sections of $\mathcal{L}$. We consider the short exact sequence $$ 0 \to \mathcal{I}\mathcal{L} \to \mathcal{L} \to  \mathcal{L}/\mathcal{I}\mathcal{L} \to 0 $$ Since the support of $\mathcal{L}/\mathcal{I}\mathcal{L}$ has dimension $0$ we see that $\mathcal{L}/\mathcal{I}\mathcal{L}$ is generated by global sections (Varieties, Lemma \ref{varieties-lemma-chi-tensor-finite}). From the short exact sequence, and the fact that $H^0(X, \mathcal{I}\mathcal{L}) = H^0(X, \mathcal{L})$ we get an injection $H^0(X, \mathcal{L}/\mathcal{I}\mathcal{L}) \to H^1(X, \mathcal{I}\mathcal{L})$. \medskip\noindent Recall that the $k$-vector space $H^1(X, \mathcal{I}\mathcal{L})$ is dual to $\Hom(\mathcal{I}\mathcal{L}, \omega_X)$. Choose $\varphi : \mathcal{I}\mathcal{L} \to \omega_X$. By Lemma \ref{lemma-vanishing-on-gorenstein} we have $H^1(X, \mathcal{L}) = 0$. Hence $$ \dim_k H^0(X, \mathcal{I}\mathcal{L}) = \dim_k H^0(X, \mathcal{L}) = \deg(\mathcal{L}) + \chi(\mathcal{O}_X) > \dim_k H^1(X, \mathcal{O}_X) = \dim_k H^0(X, \omega_X) $$ We conclude that $\varphi$ is not injective on global sections, in particular $\varphi$ is not injective. For every generic point $\eta \in X$ of an irreducible component of $X$ denote $V_\eta \subset \Hom(\mathcal{I}\mathcal{L}, \omega_X)$ the $k$-subvector space consisting of those $\varphi$ which are zero at $\eta$. Since every associated point of $\mathcal{I}\mathcal{L}$ is a generic point of $X$, the above shows that $\Hom(\mathcal{I}\mathcal{L}, \omega_X) = \bigcup V_\eta$. As $X$ has finitely many generic points and $k$ is infinite, we conclude $\Hom(\mathcal{I}\mathcal{L}, \omega_X) = V_\eta$ for some $\eta$. Let $\eta \in C \subset X$ be the corresponding irreducible component. Let $Y \subset X$ be the union of the other irreducible components of $X$. Then $Y$ is a nonempty reduced closed subscheme not equal to $X$. Let $\mathcal{J} \subset \mathcal{O}_X$ be the ideal sheaf of $Y$. Please keep in mind that the support of $\mathcal{J}$ is $C$. \medskip\noindent Let $\varphi : \mathcal{I}\mathcal{L} \to \omega_X$ be arbitrary. Since $\mathcal{J}\mathcal{I}\mathcal{L}$ has no embedded associated points (as a submodule of $\mathcal{L}$) and as $\varphi$ is zero in the generic point $\eta$ of the support of $\mathcal{J}$, we find that $\varphi$ factors as $$ \mathcal{I}\mathcal{L} \to \mathcal{I}\mathcal{L}/\mathcal{J}\mathcal{I}\mathcal{L} \to \omega_X $$ We can view $\mathcal{I}\mathcal{L}/\mathcal{J}\mathcal{I}\mathcal{L}$ as the pushforward of a coherent sheaf on $Y$ which by abuse of notation we indicate with the same symbol. Since $\omega_Y = \SheafHom(\mathcal{O}_Y, \omega_X)$ by Lemma \ref{lemma-closed-immersion-dim-1-CM} we find a factorization $$ \mathcal{I}\mathcal{L} \to \mathcal{I}\mathcal{L}/ \mathcal{J}\mathcal{I}\mathcal{L}  \xrightarrow{\varphi'} \omega_Y \to \omega_X $$ of $\varphi$. Let $\mathcal{I}' \subset \mathcal{O}_Y$ be the image of $\mathcal{I} \subset \mathcal{O}_X$. There is a surjective map $\mathcal{I}\mathcal{L}/\mathcal{J}\mathcal{I}\mathcal{L} \to \mathcal{I}'\mathcal{L}|_Y$ whose kernel is supported in closed points. Since $\omega_Y$ is a Cohen-Macaulay module on $Y$, the map $\varphi'$ factors through a map $\varphi'' : \mathcal{I}'\mathcal{L}|_Y \to \omega_Y$. Thus we have commutative diagrams $$ \vcenter{ \xymatrix{ 0 \ar[r] & \mathcal{I}\mathcal{L} \ar[r] \ar[d] & \mathcal{L} \ar[r] \ar[d] & \mathcal{L}/\mathcal{I}\mathcal{L} \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{I}'\mathcal{L}|_Y \ar[r] & \mathcal{L}|_Y \ar[r] & \mathcal{L}|_Y/\mathcal{I}'\mathcal{L}|_Y \ar[r] & 0 } } \quad\text{and}\quad \vcenter{ \xymatrix{ \mathcal{I}\mathcal{L} \ar[r]_\varphi \ar[d] & \omega_X \\ \mathcal{I}'\mathcal{L}|_Y \ar[r]^{\varphi''} & \omega_Y \ar[u] } } $$ Now we can finish the proof as follows: Since for every $\varphi$ we have a $\varphi''$ and since $\omega_X \in \textit{Coh}(\mathcal{O}_X)$ represents the functor $\mathcal{F} \mapsto \Hom_k(H^1(X, \mathcal{F}), k)$, we find that $H^1(X, \mathcal{I}\mathcal{L}) \to H^1(Y, \mathcal{I}'\mathcal{L}|_Y)$ is injective. Since the boundary $H^0(X, \mathcal{L}/\mathcal{I}\mathcal{L}) \to H^1(X, \mathcal{I}\mathcal{L})$ is injective, we conclude the composition $$ H^0(X, \mathcal{L}/\mathcal{I}\mathcal{L}) \to H^0(X, \mathcal{L}|_Y/\mathcal{I}'\mathcal{L}|_Y) \to H^1(X, \mathcal{I}'\mathcal{L}|_Y) $$ is injective. Since $\mathcal{L}/\mathcal{I}\mathcal{L} \to \mathcal{L}|_Y/\mathcal{I}'\mathcal{L}|_Y$ is a surjective map of coherent modules whose supports have dimension $0$, we see that the first map $H^0(X, \mathcal{L}/\mathcal{I}\mathcal{L}) \to H^0(X, \mathcal{L}|_Y/\mathcal{I}'\mathcal{L}|_Y)$ is surjective (and hence bijective). But by induction we have that $\mathcal{L}|_Y$ is globally generated (if $Y$ is disconnected this still works of course) and hence the boundary map $$ H^0(X, \mathcal{L}|_Y/\mathcal{I}'\mathcal{L}|_Y) \to H^1(X, \mathcal{I}'\mathcal{L}|_Y) $$ cannot be injective. This contradiction finishes the proof.