The stronger version of the Nullstellensatz states that a maximal
ideal $I$ of $R=k[x_1,\ldots,x_n]$ is the kernel of a $k$-homomorphism
from $R$ to $L$ where $L/k$ is a finite extension. Let $a_1,\ldots,a_n$
be the images of $x_1,\ldots,x_n$ under such a homomorphism.
Then $a_1$ has a minimal polynomial $m_1$ over $k$. Let
$f_1(x_1,\ldots,x_n)=m_1(x_1)$. Then $f_1\in I$
Now $a_2$ has a minimal polynomial $m_2$
over $k(a_1)$. We can write the coefficients of $m_2$ as polynomials
in $a_1$ over $k$. Doing this, and replacing $a_1$ by $x_1$ and the free
variable by $x_2$ gives a polynomial $f_2$ in $x_1$ and $x_2$. Also $f_2\in I$.
Keep going. We get a sequence of polynomials $f_i$ in $x_1,\ldots,x_i$
and it's not hard to prove these generate $I$.
ADDED: here is a proof of the statement you need (namely the square free monomial ideal $I$ is a intersection of primes generated by subsets of parameters) without using the modularity property. We will use induction on $N=$ the total numbers of times the parameters appear in the generators of $I$. For example if $I=(xy, xz)$ then $N=4$. The statement is obvious if $N=1$.
Suppose $I$ has a generator (say $f_1$) which involves at least $2$ parameters. Pick one of these parameters, say $x$ and WLOG, we can assume $I=(f_1,\cdots, f_n, g_1,\cdots,g_l) $ such that $x|f_i$ for each $i$ but $x$ does not divide any of the $g_j$s. Let $F_i=f_i/x$. We claim that:
$$ I = (I,x) \cap (I,F_1)$$
If the claim is true, we are done by applying the induction hypothesis to $(I,x)$ and $(I,F_1)$. One containment is obvious, for the other one we need to show if $xu \in (I,F_1)$ then $xu\in I$.
Write $$xu = f_2x_2 + \cdots f_nx_n + \sum g_jy_j + F_1x_1$$ which implies
$$x(u- F_2x_2 +\cdots F_nx_n) \in (g_1,\cdots, g_l, F_1) = I' $$
$I'$ has minimal generators which do not contain $x$. By induction, $I'$ is an intersection of primes generated by other parameters, so $x$ is a NZD on $R/I'$. So $(u- F_2x_2 +\cdots F_nx_n) \in I'$, and therefore $xu \in I$, as desired.
REMARK: note that for this proof to work, you only need that all subsets of the sequence (not necessarily parameters) generate prime ideals. I guess it fits with your other question.
So from the comments I will take your question as proving $J\cap (K+L) = J\cap K + J \cap L$ for parameter ideals (by which I mean ideals generated by subsets of a fixed regular s.o.p).
It will suffice to understand $I\cap J$ for two such ideals. To be precise, let $g(I)$ be the set of s.o.p generators of $I$. Let $P$ be the ideal generated by the intersection of $g(I),g(J)$, and $I', J'$ generated by $g(I)-g(P), g(J)-g(P)$. Then we need to show:
$$I \cap J = P + I'J' $$
Since $R/P$ is still regular we can kill $P$ and assume that $g(I), g(J)$ are disjoint, and we have to prove $I \cap J = IJ$. This should be an easy exercise, but a slick and very general way is invoking Tor (which shows that this is even true for $I,J$ generated by parts of a fixed regular sequence).
Best Answer
Yes, this follows from the homological characterization of regularity. Let $A=k[x_1,\dots,x_n]$ and $B=K[x_1,\dots,x_n]$, where $K$ is the algebraic closure of $k$. Then for any $A$-modules $M$ and $N$ with $M$ finitely generated, $$B\otimes \operatorname{Ext}_A^*(M,N)=\operatorname{Ext}_B^*(B\otimes M,B\otimes N).$$ Since $B$ has global dimension $n$, the right hand side vanishes for $*>n$, and hence so does $\operatorname{Ext}_A^*(M,N)$. It follows that $A$ has global dimension $n$ and is thus regular. Alternatively, there are various ways to prove directly that $A$ has finite global dimension, without reducing to the algebraically closed case (for instance, you can show that the global dimension of $R[x]$ is always one more than the global dimension of $R$).