[proofplan]
We use the ideal-sheaf exact sequence of the point $x$ and tensor it with the line bundle $M$. Since the sheaf of holomorphic sections of $M$ is locally free of rank one, tensoring preserves exactness, and the quotient term is the skyscraper sheaf whose value is the fibre $M|_x$. The long exact sequence in sheaf cohomology then shows that global sections of $M$ surject onto this fibre when $H^1(X,M\otimes_{\mathcal O_X}\mathcal I_x)$ vanishes. Applying the pointwise conclusion for every $x \in X$ is exactly the definition of global generation.
[/proofplan]
[step:Tensor the point ideal sequence by the line bundle]
Let $\mathcal Q_x$ denote the quotient sheaf $\mathcal O_X/\mathcal I_x$. The defining exact sequence of the ideal sheaf of the point $x$ is
\begin{align*}
0 \longrightarrow \mathcal I_x \longrightarrow \mathcal O_X \longrightarrow \mathcal Q_x \longrightarrow 0.
\end{align*}
Here $\mathcal Q_x$ is the skyscraper sheaf supported at $x$ with stalk $\mathbb C$ at $x$.
Since $M$ is a holomorphic line bundle, its sheaf of holomorphic sections is locally free of rank one as an $\mathcal O_X$-module. By the standard exactness of tensoring with a locally free sheaf, tensoring with $M$ over $\mathcal O_X$ preserves exactness. We obtain the short exact sequence of sheaves
\begin{align*}
0 \longrightarrow M\otimes_{\mathcal O_X}\mathcal I_x \longrightarrow M \longrightarrow M\otimes_{\mathcal O_X}\mathcal Q_x \longrightarrow 0.
\end{align*}
[/step]
[step:Identify the quotient sheaf with the fibre at $x$]
Let $M_{x,\mathrm{st}}$ denote the stalk at $x$ of the sheaf of holomorphic sections of $M$, and let $M|_x$ denote the fibre of the holomorphic line bundle at $x$. The sheaf $M\otimes_{\mathcal O_X}\mathcal Q_x$ is supported only at $x$. Its stalk at $x$ is
\begin{align*}
(M\otimes_{\mathcal O_X}\mathcal Q_x)_x
\cong M_{x,\mathrm{st}}\otimes_{\mathcal O_{X,x}}(\mathcal O_{X,x}/\mathcal I_{x,x}).
\end{align*}
Since $\mathcal I_{x,x}$ is the maximal ideal of holomorphic germs vanishing at $x$, the quotient $\mathcal O_{X,x}/\mathcal I_{x,x}$ is $\mathbb C$. The standard fibre construction for a holomorphic vector bundle gives
\begin{align*}
M_{x,\mathrm{st}}\otimes_{\mathcal O_{X,x}}\mathbb C \cong M|_x.
\end{align*}
Hence
\begin{align*}
(M\otimes_{\mathcal O_X}\mathcal Q_x)_x \cong M|_x.
\end{align*}
Because the sheaf is supported at the single point $x$, taking global sections gives
\begin{align*}
H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)\cong M|_x.
\end{align*}
Under this identification, the map
\begin{align*}
H^0(X,M)\longrightarrow H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)
\end{align*}
is precisely the evaluation map $\operatorname{ev}_x:H^0(X,M)\to M|_x$.
[/step]
[step:Use the long exact sequence to force surjectivity]
Apply the standard long exact sequence in sheaf cohomology to
\begin{align*}
0 \longrightarrow M\otimes_{\mathcal O_X}\mathcal I_x \longrightarrow M \longrightarrow M\otimes_{\mathcal O_X}\mathcal Q_x \longrightarrow 0.
\end{align*}
The hypothesis needed for this external result is exactly that the displayed sequence is a short exact sequence of sheaves of abelian groups on $X$, which was proved after tensoring the point ideal sequence by the locally free $\mathcal O_X$-module $M$. It gives, in low degrees, the exact sequence
\begin{align*}
H^0(X,M) \longrightarrow H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x) \xrightarrow{\delta_x} H^1(X,M\otimes_{\mathcal O_X}\mathcal I_x).
\end{align*}
Here $\delta_x$ denotes the connecting homomorphism.
By hypothesis,
\begin{align*}
H^1(X,M\otimes_{\mathcal O_X}\mathcal I_x)=0.
\end{align*}
Therefore $\delta_x$ is the zero map into the zero [vector space](/page/Vector%20Space). Exactness says that the image of
\begin{align*}
H^0(X,M)\longrightarrow H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)
\end{align*}
is the kernel of $\delta_x$. Since $\ker\delta_x$ is all of $H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)$, this map is surjective. By the fibre identification from the previous step, $\operatorname{ev}_x:H^0(X,M)\to M|_x$ is surjective.
[guided]
The goal is to prove that every vector in the fibre $M|_x$ is the value at $x$ of some global holomorphic section of $M$. Let $M_{x,\mathrm{st}}$ denote the stalk at $x$ of the sheaf of holomorphic sections of $M$. The quotient sheaf $\mathcal Q_x=\mathcal O_X/\mathcal I_x$ is supported at $x$, and its stalk is $\mathcal O_{X,x}/\mathcal I_{x,x}\cong\mathbb C$. Therefore
\begin{align*}
(M\otimes_{\mathcal O_X}\mathcal Q_x)_x
\cong M_{x,\mathrm{st}}\otimes_{\mathcal O_{X,x}}\mathbb C
\cong M|_x.
\end{align*}
Since $M\otimes_{\mathcal O_X}\mathcal Q_x$ is supported only at $x$, global sections are exactly the value of the stalk at $x$, so
\begin{align*}
H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)\cong M|_x.
\end{align*}
This is why the quotient sheaf isolates the fibre.
Now apply the standard long exact sequence in sheaf cohomology to the short exact sequence
\begin{align*}
0 \longrightarrow M\otimes_{\mathcal O_X}\mathcal I_x \longrightarrow M \longrightarrow M\otimes_{\mathcal O_X}\mathcal Q_x \longrightarrow 0.
\end{align*}
The required hypothesis is that this is a short exact sequence of sheaves of abelian groups on $X$, and this holds by the standard exactness of tensoring with a locally free sheaf, since $M$ is a locally free $\mathcal O_X$-module of rank one. The long exact sequence gives the exact segment
\begin{align*}
H^0(X,M) \longrightarrow H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x) \xrightarrow{\delta_x} H^1(X,M\otimes_{\mathcal O_X}\mathcal I_x).
\end{align*}
The symbol $\delta_x$ denotes the connecting homomorphism attached to this short exact sequence.
Exactness at the middle term means
\begin{align*}
\operatorname{im}\left(H^0(X,M)\to H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)\right)=\ker \delta_x.
\end{align*}
The vanishing hypothesis says
\begin{align*}
H^1(X,M\otimes_{\mathcal O_X}\mathcal I_x)=0.
\end{align*}
Therefore $\delta_x$ maps into the zero vector space, so every element of $H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)$ lies in $\ker\delta_x$. Hence
\begin{align*}
\ker\delta_x=H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x).
\end{align*}
Combining this equality with exactness shows that
\begin{align*}
H^0(X,M)\longrightarrow H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)
\end{align*}
is surjective. Under the identification
\begin{align*}
H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)\cong M|_x,
\end{align*}
this is exactly the evaluation map $\operatorname{ev}_x:H^0(X,M)\to M|_x$. Thus every element of the fibre $M|_x$ is the value at $x$ of a global section of $M$.
[/guided]
[/step]
[step:Conclude global generation from pointwise surjectivity]
Assume now that
\begin{align*}
H^1(X,M\otimes_{\mathcal O_X}\mathcal I_x)=0
\end{align*}
for every $x \in X$. Applying the pointwise result just proved to each $x$ shows that
\begin{align*}
\operatorname{ev}_x:H^0(X,M)\longrightarrow M|_x
\end{align*}
is surjective for every $x \in X$. This is precisely the definition that the holomorphic line bundle $M$ is globally generated. Hence $M$ is globally generated.
[/step]
