I think I might be to blame for this question. It looks very similar to something I once wrote, with the same gap.

The result is true, but the approach described will not work. We have to choose the $f_n$ with more care.

(Indeed, suppose $x$ is outside the linear span of your $x_n$. Note that $E$ is closed, hence so is $E+x$. You could thus choose the functionals $f_n$ perversely using Hahn-Banach so that they all vanish on $E +x$; if it does in fact turn out to be the case that $\mu(E)=1$ then you'll still have $q(f_n, f_n)=0$. Then $x$ is in $\bigcap_n \ker f_n$ but not in $E$.)

I think the following should work instead. Let $F = \{f \in X^* : q(f,f) = 0\}$. Since $X$ is separable, the unit ball $B^*$ of $X^*$ is weak-* separable metrizable, hence so is $B^* \cap F$. So pick your sequence $\{f_n\}$ to be weak-* dense in $B^* \cap F$. Suppose $f_n(x) = 0$ for all $n$ and take any $f \in F$, renormalized so that $\|f\| \le 1$. Pass to a subsequence so that $f_n \to f$ weak-*. Then $0 = f_n(x) \to f(x)$. Since $f$ was arbitrary we have $x \in E$.

I think I might be to blame for this question. It looks very similar to something I once wrote, with the same gap. If so, sorry!

The result is true, but the approach described will not work. We have to choose the $f_n$ with more care.

(Indeed, suppose $x$ is outside the linear span of your $x_n$. Note that $E$ is closed, hence so is $E+x$. You could thus choose the functionals $f_n$ perversely using Hahn-Banach so that they all vanish on $E +x$; if it does in fact turn out to be the case that $\mu(E)=1$ then you'll still have $q(f_n, f_n)=0$. Then $x$ is in $\bigcap_n \ker f_n$ but not in $E$.)

I think the following should work instead. Let $F = \{f \in X^* : q(f,f) = 0\}$. Since $X$ is separable, the unit ball $B^*$ of $X^*$ is weak-* separable metrizable, hence so is $B^* \cap F$. So pick your sequence $\{f_n\}$ to be weak-* dense in $B^* \cap F$. Suppose $f_n(x) = 0$ for all $n$ and take any $f \in F$, renormalized so that $\|f\| \le 1$. Pass to a subsequence so that $f_n \to f$ weak-*. Then $0 = f_n(x) \to f(x)$. Since $f$ was arbitrary we have $x \in E$.