Consider $k$ a field. Identify $\otimes_{i\leq n}k[x]=k[x_1,\dots, x_n]$ where tensor is over $k$ as algebra.
I want to see the following holds. Let $z=(v_1,\dots, v_n)\in k^n, z'=(v'_1,\dots, v'_n)\in k^n$. Assume $z\neq z'$. I want to find $f$ below invariant under $S_n$ action.(i.e. $S_n$ permutes $v_i$'s above.)
"... Then we may assume that $v_1\neq v'_1$. Then there is an $f_1\in k[x]$ s.t. $f_1(v_1)=0$ and $f_1(v'_1)=1$. Now choose $f_i\in k[x],2\leq i\leq n$ s.t. $f_i(v_i)=f_i(v'_i)=1$. Let $f=\otimes f_i\in \otimes_i k[x]=k[x_1,\dots, x_n]$. Clearly $f$ is invariant and $f(z)=0,f(z')=1$."
$\textbf{Q:}$ Why is this $f$ invariant? Naive choice of $f_1$ is linear in $x_1$ and $f_i$ quadratic in $x_i$ for $2\leq i\leq n$. There is no way this is symmetric as it is not even homogeneous of same degree. I do not even believe symmetrization of above $f$ will necessarily guarantee $f(z)=0,f(z')=1$ in general.
The following is my thought for producing such $f$ for $char(k)\neq 0$ and $k$ need not be algebraically closed. It suffices to product a polynomial $f$ s.t. $f(z)=0$ and $f(z')\neq 0$. Consider $f=\sum_ia_i\sum_j(x_j-v_j)^i$ which is clearly invariant under $S_n$. Clearly $f(z)=0$. Now it suffices to choose $a_i$ s.t. $f(v'_j)\neq 0$. Then I need to assume $char(k)\neq 0$ at least or I can't conclude such $f$ exists by my construction. Assume for contradiction such $f$ does not exist.(i.e. $\sum_j(v'_j-v_j)^i=0$ for all $i\geq 1$. Then I can use induction to conclude that $v'_j=v_j$.(This is equivalent to show $\sum_j a_j^i=0$ for $i\geq 1$. One can assume $a_j\neq 0$ for all $j$ by induction. Then use $i$ th equation to eliminate $i+1$ equation's $a_1^{i+1}$ part. This will yield another set of equations. Then eliminate $a_2^i(a_2-a_1)$ part. This procedure terminates with $a_n\prod_{j\neq n}(a_n-a_j)=0$. Say $a_n=a_k$. Now $S_n$ symmetry yields $a_i=a_n$ for all $i$. Thus $a_i=0$ for all $i$ by $char(k)\neq 0$.) Thus we conclude there is such a polynomial $f$.