The the image of the unit ball in X is weak-* dense in the unit ball of X**

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

$$Q'(B_x) = B_{x^{\ast\ast}}.$$