$E|X-m|$ is minimised at $m$=median

I'll denote your expression by $R(m)$; then

$$R'(m) = 2 F(m) + 2 m f(m) - 1 + 0 - 2 m f(m) = 2 F(m) - 1$$

where I used the FTC on the last term. So $R'(m)=0$ if and only if $F(m)=1/2$, as desired. That gives you a critical point. Now it is a minimum provided that $f(m)>0$, because

$$R''(m) = 2 f(m).$$

If it is a minimum then it is a global minimum, because $R$ is convex. If it is not a minimum then, because $R'' \geq 0$ globally, $R$ is actually locally constant near the candidate for the median. This implies that there is not a unique median. One important case where this occurs is with a uniform distribution on a discrete set with an even number of elements. In this case, supposing the elements are listed in order as $\{ x_i \}_{i=1}^{2k}$, any number in $[x_k,x_{k+1})$ will be a median.