Two numbers are independently picked from [0,1], what's the probability that the two picks differ by less than 0.5? So using a graphical method, mapping the two sample spaces along X and Y axes in a unit square, I graphically calculate the area relevant to this probability to be $\frac{3}{4}$
But, using the Random Variable $X$ where the events are the differences between the two picks (min difference is $0-1=-1$ and similarly max is $1$)
This yields a Probability Density Function where there is a horizontal line of magnitude $\frac{1}{b-a} = \frac{1}{2}$ over $[-1,1]$ but is $0$ elsewhere.
Now taking the integral over $[-0.5,0.5]$ to find $P[-0.5 <= x < 0.5]$ yields $\frac{1}{2}$
Why is this? I believe that there is an incorrect of the $PDF$ somewhere. 

If I understand your post correctly, your PDF is incorrect.  The PDF of the (absolute value) difference $Z$ between two variables $X$ and $Y$, both uniformly drawn from the interval $[0, 1]$, is in fact
$$
f_Z(z) = \begin{cases}
2-2z & 0 \leq z \leq 1 \\
0 & \text{otherwise}
\end{cases}
$$
This can be obtained as follows: The CDF (cumulative distribution function) of $Z$ for $0 < z < 1$ is
\begin{align}
F_Z(z) & = P(Z < z) \\
       & = P(X-z < Y < X+z) \\
       & = \int_{x=0}^1 2z \, dx
         - \int_{x=0}^z (z-x) \, dx
         - \int_{x=1-z}^1 (z-1+x) \, dx \\
       & = 2z-\left(z^2-\frac{z^2}{2}\right)-\left(z^2-z+z-\frac{z^2}{2}\right) \\
       & = 2z-z^2
\end{align}
so
$$
f_Z(z) = \frac{d}{dz} F_Z(z) = 2-2z
$$
in that same interval.
