Why that a quasi-projective variety being isomorphic to a projective variety must itself be projective as well

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On page 54 in the Book An Invitation to Algebraic Geometry, the author said that

A quasi-projective variety in $\mathbb{P}^n$ is isomorphic to a Zariski-closed subset of some $\mathbb{P}^m$ (i.e. a projective variety) if and only if it already forms a Zariski-closed subset of $\mathbb{P}^n$.

Can anyone provide a proof?

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I think Hoot's answer is the correct one.

If $X/k$ is projective, then any locally closed embedding $X\hookrightarrow \mathbb{P}^n_k$ must be a closed embedding. Indeed, since $X/k$ is proper, and $\mathbb{P}^n_k$ separated, we know from the Cancellation Lemma (see Vakil) that $X\to \mathbb{P}^n_k$ is proper, and in particular, closed. But, any closed embedding with closed image is a closed embedding.