Given the quotient ring $S/I$, where $S=K[x_1,...,x_n]$ is a polynomial ring and $I$ is an ideal. I need to show that for any monomial $x^u:=x_1^{u_1}\cdots x_n^{u_n}$, if the monomial is not inside $\operatorname{in}_<(I):=\langle\{\operatorname{in}_<(f)\mid f\in I\}\rangle$, then the monomial is part of a $K$-basis for the vector space $S/I$.
I've already shown that $S/I$ is a vector space.
I assumed that $x^v\notin\operatorname{in}_<(I)$, which implies $x^v\notin I$. Since $S/I=\{s+I|s\in S\}$, and $x^v\in S$, $0\in I$ (can I say that $0\in I$, or is that not a general case with ideals), then $(x^v +0)\in S/I$. So for the monomials $x^{v_1},...,x^{v_m}$ with the same property as $x^v$, I need to show that they are all linearly independent in $S/I$ and span $S/I$. However, I'm struggling on proving this.
Consider the set of $x^{v}$ that are not in $\operatorname{in}_{<}(I)$.
Linear independence is easy - if a linear combination $F$ of the set is zero in $S/I$ then when considered as an element of $S$ we have $F\in I$ so one of them is in $\operatorname{in}_{<}(I)$, contradiction.
Spanning is harder. Consider the set $H$ of non zero $f\in S$ that are non zero in $S/I$ and are not a linear combination of our set. Then in each $f\in H$ there is some maximal term $x^u$ w.r.t < which is not in our set, so is in $\operatorname{in}_{<}(I)$. If $H$ is non empty we can take the $f$ which has the minimal one of these. We know then that there is some $g\in I$ with $\operatorname{in}_{<}(g)=x^u$. Taking the correct multiple of $g$ from $f$ gives polynomial in $H$ which has a smaller maximal element than $f$, contradiction. So $H$ is empty, so our set spans $S/I$.