The correct answer is $\boxed{\text{B) }0}$.
The z-transform of a sequence $x[n]$ is defined as
$$X[z] = \sum_{n=-\infty}^{\infty} x[n] z^{-n}$$
The region of convergence (ROC) of $X[z]$ is the set of all values of $z$ for which the series converges.
In this case, we are given that the ROC of $X[z]$ includes the unit circle. This means that for any value of $z$ on the unit circle, the series
$$\sum_{n=-\infty}^{\infty} x[n] z^{-n}$$
converges.
We can use this information to determine the value of $x[0]$. The coefficient of $z^{-0}$ in the series is $x[0]$, so we have
$$x[0] = \lim_{z \to 1} X[z]$$
Substituting the given value of $X[z]$, we get
$$x[0] = \lim_{z \to 1} \frac{0.5}{1 – 2z^{-1}} = \lim_{z \to 1} \frac{0.5}{1 – 2 \frac{1}{z}} = \lim_{z \to 1} \frac{0.5z}{z – 2}$$
The limit of the expression on the right-hand side as $z$ approaches 1 is 0, since the numerator and denominator both approach 0. Therefore, the value of $x[0]$ is 0.