This is Vakil 18.6 M, self-study.
We are to show that an integral degree $d$ curve $C$ in $\mathbb P^N_k$ (with $N \geq d$ and $k$ WLOG algebraically closed) is contained in a "linear $\mathbb P^d_k \subset \mathbb P^N_k$."
So, far I see that if you take a hyperplane $H$ in $\mathbb P^N_k$ intersecting $C$ but not containing it, then by Bezout's Theorem (18.6 K in Vakil),
$$\deg C \cap H = \deg C \deg H = d$$
where the degree indicates the leading coefficient of the Hilbert polynomial times $n!$, where $n$ is the dimension of the space. The problem is that I do not see what this tells us geometrically. I can see the leading coefficient of the Hilbert polynomial of $C \cap H$ must have been $d$ to begin with, but this seems purely formal.
I have studied both answers here, but we have not covered general position (the content of the first answer), and I do not follow the conclusion of the second answer, in addition to the fact that it does not seem to use Bezout's Theorem, at least as we have covered it. I suspect neither answer is what Vakil is looking for, and that I am simply missing some geometric realization about degree that would quickly convert what I have done thus far into an answer.
By "a linear $\mathbb P^d_k$" I assume Vakil means a linear space of dimension $d$ in $\mathbb P^N_k$, meaning a closed subscheme cut out by $N – d$ homogeneous linearly independent linear polynomials.
Best Answer
Continued from the comments - we established there that if $C\not\subset H$, then $|C\cap H|\leq d$. If you can only ever get at most $d$ points with a proper intersection, what happens when you pick a hyperplane through $d+1$ points on $C$? Full details under the spoiler.