With the same notations in you question: Notice that if $x_i^*(x) = 0$ for all $i$, then $x \in U$, and therefore the intersection of the kernels $\bigcap_{i=1}^n \mathrm{ker}(x_i^*)$ is in $U$. Since the codimension of $\mathrm{ker}(x^*_i)$ is at most $1$, then the intersection has codimension at most $n$ (exercise: prove this). But since $X$ is infinite dimensional, this means the intersection has an infinite dimension, and in particular contains a line. Since any line going through $0$ intersects $S$, then $U$ intersects $S$.
The same argument can be applied to any point in $B$ (any line going through a point in $B$ intersects $S$), and since you've proved the other inclusion, the weak closure of $S$ is $B$.
You are sort of two inches from the finish line.
We know that $Q'(B_x)$ is a convex, balanced, and (weak*) closed subset of $X^{\ast\ast}$ (that is contained in $B_{x^{\ast\ast}}$).
By (one of) the Hahn-Banach theorem(s), for every $\pi \notin Q'(B_x)$, there is a continuous linear form $\lambda$ on $X^{\ast\ast}$ (endowed with the weak* topology) with $\lvert\lambda(\eta)\rvert \leqslant 1$ for all $\eta \in Q'(B_x)$, and $\lvert\lambda(\pi)\rvert > 1$.
The difference to what you have is (apart from specifying $\alpha = 1$) the strict inequality for $\lvert\lambda(\pi)\rvert$, where you only had a weak inequality. That difference is crucial, however.
Now, by definition of the weak* topology on $X^{\ast\ast}$, the space of continuous linear forms is $X^\ast$.
Then $\lvert \lambda(\eta)\rvert \leqslant 1$ for all $\eta \in Q'(B_x)$ implies $\lvert\lambda(x)\rvert \leqslant 1$ for all $x\in B_x$, which means $\lVert \lambda\rVert \leqslant 1$.
But then
$$1 < \lvert \pi(\lambda)\rvert \leqslant \lVert \pi\rVert\cdot\lVert\lambda\rVert \leqslant \lVert\pi\rVert.$$
So $\pi \notin B_{x^{\ast\ast}}$. That holds for all $\pi \notin Q'(B_x)$, hence
$$Q'(B_x) = B_{x^{\ast\ast}}.$$
Best Answer
It's basic point-set topology. The closure of a set $X$ (in a certain topology, such as the weak topology here) is the intersection of all closed sets containing $X$, or equivalently the complement of the union of all open sets that don't intersect $X$.
If a point $a$ is not in the closure of $X$, that means there's an open set containing $a$ (namely the complement of the closure of $X$) that doesn't intersect $X$. By the contrapositive, if every open set containing $a$ intersects $X$, then $a$ is in the closure of $X$. Done.