(See statement and proof attached below.) In his proof of Corollary 1.9, he states that $\mathfrak{m}_p \supset I(X)$ iff $p\in X$. I can see one direction, namely if $p\in X$ then $\mathfrak{m}_p \supset I(X)$. Why is the other direction true? I think it has to do with the fact that $I(X)$ is a radical ideal, but I'm not sure how to leverage this fact. Thanks!
Corollary 1.9. Let $k$ be a field. For each $p=(a_1,\ldots,a_r)\in\mathbf{A}^r(k)$ the ideal $\mathfrak{m}_p:=(x_1-a_1,\ldots,x_n-a_n)\subset k[a_1,\ldots,x_r]$ is a maximal ideal. If $k$ is algebraically closed and $X\subset\mathbf{A}^r(k)$ is an algebraic set, then every maximal ideal of $A(X)$ is of the form $\mathfrak{m}_p/I(X)$ for some $p\in X$. In particular, the points of $X$ are in one-to-one correspondence with the maximal ideals of the ring $A(X)$.
Proof. It is clear that $k[x_1,\ldots,x_r]/\mathfrak{m}_p$, so $\mathfrak{m}_p$, is a maximal ideal. The natural map $k[x_1,\ldots,x_r]\rightarrow k[x_1,\ldots,x_r]/\mathfrak{m}_p=k$ may be described as evaluation at $p$. Thus $\mathfrak{m}_p\supset I(X)$ iff $p\in X$. Since the maximal ideals of $A(X)$ are the maximal ideals of $S:=[x_1,\ldots,x_r]$ containing $I(X)$, taken modulo $I(X)$, it only remains to show that every maximal ideal of $S$ has the form $\mathfrak{m}_p$ for some $p$.
Suppose $\mathfrak{n}$ is a maximal ideal of $S$. By Theorem 4.19 applied with $R=k$, $S/\mathfrak{n}$ is algebraic over $k/(\mathfrak{n}\cap k)=k$. Since $k$ is algebraically closed, $S/\mathfrak{n}=k$. Let $a_i$ be the image of $x_i$ under the map $S\rightarrow S/\mathfrak{n}=k$, and let $p=(a_1,\ldots,a_r)$. It follows that $\mathfrak{m}_p$ is contained in $\mathfrak{n}$. Since $\mathfrak{m}_p$ is maximal, $\mathfrak{m}_p=\mathfrak{n}$. $\qquad\square$
