Is an algebraic set, all of whose irreducible components are projective varieties, projective too?

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Let $X$ be an algebraic set (that is, a non-necessarily irreducible algebraic variety) over a field $k$. In the language of schemes, let us say that $X$ is a reduced separated scheme of finite type over $\mathrm{Spec}(k)$. Then $X$ has a finite number of irreducible components which we denote by $X_i$ for $1\leq i \leq r$.

Each of the $X_i$, with the reduced scheme structure, is a closed subscheme of $X$ and a variety over $k$. We assume that they are furthermore all projective.

Does it follow that $X$ is projective too ?

Sure thing, $X$ should be a proper scheme, as the universal closeness of the structure morphism $X\to \mathrm{Spec}(k)$ should follow from that of the irreducible components, since they are in finite number.
What about quasi-projectiveness though ?