**Proof.**
By Lemma 53.19.16 we find that $X$ is Gorenstein, i.e., $\omega _ X$ is an invertible $\mathcal{O}_ X$-module.

If the genus of $X$ is zero, then $\deg (\omega _ X) < 0$, hence if $X$ has more than one irreducible component, we get a contradiction with Lemma 53.22.4. In the irreducible case we see that $X$ is isomorphic to an irreducible plane conic and $\omega _ X^{\otimes -1}$ is very ample by Lemma 53.10.3.

If the genus of $X$ is $1$, then $\omega _ X$ has a global section and $\deg (\omega _ X|_ C) = 0$ for all irreducible components. Namely, $\deg (\omega _ X|_ C) \geq 0$ for all irreducible components $C$ by Lemma 53.22.4, the sum of these numbers is $0$ by Lemma 53.8.3, and we can apply Varieties, Lemma 33.44.6. Then $\omega _ X \cong \mathcal{O}_ X$ by Varieties, Lemma 33.44.13.

Assume the genus $g$ of $X$ is greater than or equal to $2$. If $X$ is irreducible, then we are done by Lemma 53.21.3. Assume $X$ reducible. By Lemma 53.22.4 the inequalities of Lemma 53.21.7 hold for every $Y \subset X$ as in the statement, except for $Y = X$. Analyzing the proof of Lemma 53.21.7 we see that (in the reducible case) the only inequality used for $Y = X$ are

\[ \deg (\omega _ X^{\otimes m}) > -2 \chi (\mathcal{O}_ X) \quad \text{and}\quad \deg (\omega _ X^{\otimes m}) + \chi (\mathcal{O}_ X) > \dim _ k H^1(X, \mathcal{O}_ X) \]

Since these both hold under the assumption $g \geq 2$ and $m \geq 2$ we win.
$\square$

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