Proof.
Choose basis s_0, \ldots , s_ n of H^0(X, \mathcal{L}^{\otimes 2}) over k. By property (1) we see that \mathcal{L}^{\otimes 2} is globally generated and we get a morphism
\varphi _{\mathcal{L}^{\otimes 2}, (s_0, \ldots , s_ n)} : X \longrightarrow \mathbf{P}^ n_ k
See Constructions, Section 27.13. The lemma asserts that this morphism is a closed immersion. To check this we may replace k by its algebraic closure, see Descent, Lemma 35.23.19. Thus we may assume k is algebraically closed.
Assume k is algebraically closed. For each generic point \eta _ i \in X let V_ i \subset H^0(X, \mathcal{L}) be the k-subvector space of sections vanishing at \eta _ i. Since \mathcal{L} is globally generated, we see that V_ i \not= H^0(X, \mathcal{L}). Since X has only a finite number of irreducible components and k is infinite, we can find s \in H^0(X, \mathcal{L}) nonvanishing at \eta _ i for all i. Then s is a regular section of \mathcal{L} (because X is Cohen-Macaulay by Lemma 53.6.1 and hence \mathcal{L} has no embedded associated points).
In particular, all of the statements given in the proof of Lemma 53.7.1 hold with this s. Moreover, as \mathcal{L} is globally generated, we can find a global section t \in H^0(X, \mathcal{L}) such that t|_ Z is nonvanishing (argue as above using the finite number of points of Z). Then in the proof of Lemma 53.7.1 we can use t to see that additionally the multiplication map
\mu _ n : H^0(X, \mathcal{L}) \otimes _ k H^0(X, \mathcal{L}^{\otimes 2}) \longrightarrow H^0(X, \mathcal{L}^{\otimes 3})
is surjective. Thus
S = \bigoplus \nolimits _{n \geq 0} H^0(X, \mathcal{L}^{\otimes n})
is generated in degrees 0, 1, 2 over k. Arguing as in the proof of Lemma 53.7.1 we find that S^{(2)} = \bigoplus _{n} S_{2n} is generated in degree 1. This means that \mathcal{L}^{\otimes 2} is very ample as desired. Some details omitted.
\square
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