Proof (for future reference) : Let $\mathcal{I} \subset R[x]$ be an ideal. Start with a polynomial $f_i \in \mathcal{I}$ of minimal degree. Inductively, pick polynomial $f_{n+1} \in \mathcal{I}$ of minimal degree which is not in $(f_1, f_2, \cdots, f_n)$. If $(f_1, f_2, \cdots, f_i) = \mathcal{I}$, we stop and there is nothing to prove.

Let $a_i$ be initial coefficients of $f_i$. $\mathcal{J} = (a_1, a_2, \cdots ) \subset R$ forms an ideal. By Noetherianness, this ideal is finitely generated, hence can be generated by finitely many of the $a_i$'s. Let $m$ be the smallest integer such that $a_1, a_2, \cdots, a_m$ generates $\mathcal{J}$. We claim that $(f_1, \cdots, f_m) = \mathcal{I}$.

Assume it doesn't. Then we can similarly build $f_{m+1}$ using our algorithm mentioned before. $a_{m+1}$, the initial coefficient of $f_{m+1}$, can be written as a linear combination $\sum_{j = 1}^m c_j a_j$, as $a_{m+1} \in \mathcal{J} = (a_1, \cdots, a_m)$. As we have the following inequality$\deg f_{m+1} \geq \max \{\deg f_i\}_{i = 1}^m$ by construction of $f_{m+1}$, the polynomial $g(x) = \sum_{i = 1}^m c_i f_i x^{\deg f_{m+1} - \deg f_i}$ has same initial coefficient and degree as $f_{m+1}$

Thus, $f_{m+1} - g \in \mathcal{I}$ has degree strictly less than $f_{m+1}$, but is not an element of $(f_1, \cdots, f_m)$. This contradicts the minimality of the degree of $f_{m+1}$. Hence, $\mathcal{I} = (f_1, \cdots, f_m)$ is finitely generated.

This is a charming proof of the theorem as given in Eisenbud.