[proofplan] We use the short exact sequence defining the ideal sheaf of the two-point subscheme $\{x,y\}$. Tensoring by the line bundle $M$ preserves exactness, and the quotient becomes the skyscraper sheaf with fibres $M_x$ and $M_y$. The long exact sequence in sheaf cohomology shows that the only obstruction to prescribing arbitrary values in $M_x\oplus M_y$ lies in $H^1(X,M\otimes\mathcal I_{x,y})$, which vanishes by hypothesis. [/proofplan]

[step:Tensor the two-point ideal sequence by the line bundle $M$]Let $i_x:\{x\}\hookrightarrow X$ and $i_y:\{y\}\hookrightarrow X$ denote the inclusion maps. Let $\mathbb{C}_x$ and $\mathbb{C}_y$ denote copies of the one-dimensional complex [vector space](/page/Vector%20Space) $\mathbb{C}$ assigned to the points $x$ and $y$, respectively, so that $i_{x*}\mathbb{C}_x$ and $i_{y*}\mathbb{C}_y$ are skyscraper sheaves on $X$. By definition of $\mathcal{I}_{x,y}$, there is a short exact sequence of coherent sheaves \begin{align*} 0 \longrightarrow \mathcal{I}_{x,y} \longrightarrow \mathcal{O}_X \longrightarrow i_{x*}\mathbb{C}_x \oplus i_{y*}\mathbb{C}_y \longrightarrow 0, \end{align*} where the map $\mathcal{O}_X \to i_{x*}\mathbb{C}_x \oplus i_{y*}\mathbb{C}_y$ sends a [holomorphic function](/page/Holomorphic%20Function) germ to its pair of values at $x$ and $y$. Since $M$ is a holomorphic line bundle, the sheaf of holomorphic sections of $M$ is locally free of rank $1$, so tensoring by $M$ is exact. Therefore we obtain a short exact sequence \begin{align*} 0 \longrightarrow M\otimes \mathcal{I}_{x,y} \longrightarrow M \longrightarrow M\otimes (i_{x*}\mathbb{C}_x \oplus i_{y*}\mathbb{C}_y) \longrightarrow 0. \end{align*} The quotient sheaf is canonically identified with the skyscraper sheaf whose stalks at $x$ and $y$ are $M_x$ and $M_y$, respectively. Thus the sequence becomes \begin{align*} 0 \longrightarrow M\otimes \mathcal{I}_{x,y} \longrightarrow M \longrightarrow i_{x*}M_x \oplus i_{y*}M_y \longrightarrow 0. \end{align*}[/step]

[guided]Let us unpack the exact sequence carefully. The sheaves $i_{x*}\mathbb{C}_x$ and $i_{y*}\mathbb{C}_y$ are the skyscraper sheaves whose only nonzero stalks are copies of $\mathbb{C}$ at $x$ and $y$, respectively. The sheaf $\mathcal{I}_{x,y}$ consists of those local holomorphic functions that vanish at both points $x$ and $y$. Hence the quotient $\mathcal{O}_X/\mathcal{I}_{x,y}$ records only the two values of a function at $x$ and at $y$, which gives the skyscraper sheaf \begin{align*} \mathcal{O}_X/\mathcal{I}_{x,y}\cong i_{x*}\mathbb{C}_x \oplus i_{y*}\mathbb{C}_y. \end{align*} This is the sheaf-theoretic form of evaluating a function at two points. Since $M$ is a line bundle, its sheaf of sections is locally free of rank $1$. Tensoring a short exact sequence by a locally free sheaf preserves exactness, so tensoring the defining sequence by $M$ gives \begin{align*} 0 \longrightarrow M\otimes \mathcal{I}_{x,y} \longrightarrow M \longrightarrow M\otimes (i_{x*}\mathbb{C}_x \oplus i_{y*}\mathbb{C}_y) \longrightarrow 0. \end{align*} The [tensor product](/page/Tensor%20Product) on the quotient simply replaces the scalar value at each point by the corresponding value in the fibre of $M$. Therefore \begin{align*} M\otimes (i_{x*}\mathbb{C}_x \oplus i_{y*}\mathbb{C}_y)\cong i_{x*}M_x \oplus i_{y*}M_y. \end{align*} Thus we have the exact sequence \begin{align*} 0 \longrightarrow M\otimes \mathcal{I}_{x,y} \longrightarrow M \longrightarrow i_{x*}M_x \oplus i_{y*}M_y \longrightarrow 0. \end{align*} The last map is precisely evaluation of a local section of $M$ at the two fibres.[/guided]

[step:Use the vanishing hypothesis to make the evaluation map surjective] The long exact sequence in sheaf cohomology associated to the short exact sequence gives the exact segment \begin{align*} H^0(X,M) \xrightarrow{\operatorname{ev}_{x,y}} H^0(X,i_{x*}M_x \oplus i_{y*}M_y) \longrightarrow H^1(X,M\otimes \mathcal{I}_{x,y}). \end{align*} Since $i_{x*}M_x$ and $i_{y*}M_y$ are skyscraper sheaves, their global sections are their fibres, so \begin{align*} H^0(X,i_{x*}M_x \oplus i_{y*}M_y)\cong M_x\oplus M_y. \end{align*} Under this identification, the first arrow is the evaluation map \begin{align*} \operatorname{ev}_{x,y}:H^0(X,M)\longrightarrow M_x\oplus M_y. \end{align*} The hypothesis says that \begin{align*} H^1(X,M\otimes \mathcal{I}_{x,y})=0. \end{align*} Let $\operatorname{im}(\operatorname{ev}_{x,y})$ denote the image subspace of the evaluation map, and let $\ker\left(M_x\oplus M_y \longrightarrow H^1(X,M\otimes \mathcal{I}_{x,y})\right)$ denote the kernel of the connecting homomorphism. Exactness therefore gives \begin{align*} \operatorname{im}(\operatorname{ev}_{x,y})=\ker\left(M_x\oplus M_y \longrightarrow H^1(X,M\otimes \mathcal{I}_{x,y})\right)=M_x\oplus M_y. \end{align*} Hence $\operatorname{ev}_{x,y}$ is surjective. [/step]

[step:Translate surjectivity into separation of the two points] Because $M_y$ is a one-dimensional complex vector space, choose a nonzero vector $v_y\in M_y$. Since $\operatorname{ev}_{x,y}$ is surjective, there exists a section $s\in H^0(X,M)$ such that \begin{align*} \operatorname{ev}_{x,y}(s)=(0,v_y). \end{align*} Equivalently, $s(x)=0$ and $s(y)=v_y\ne 0$. Thus the section $s$ distinguishes the two points in the complete linear system $|M|$. Therefore $|M|$ separates $x$ and $y$. [/step]