The right formula is 
$$\Delta X \Delta P \geq  h/4\pi$$
where $P$ is the momentum which is approximatively  $mv$ only for small velocities $v$ when compared with $c$. Otherwise you have to use the relativistic expression
$$P = mv/ \sqrt{1-v^2/c^2}.$$
If $\Delta X$ is small, then $\Delta P$ is large but, according to the formula above,  the speed remains of the order of $c$ at most. That is  because, in the formula  above, $P\to +\infty$ corresponds to $v\to c$.

With some details, solving the above identity for $v$, we have
$$v = \frac{P}{m \sqrt{1+ P^2/m^2c^2}}\:,$$ so that
$$v\pm \Delta v = \frac{P\pm \Delta P}{m \sqrt{1+ (P\pm \Delta P)^2/m^2c^2}}.$$
We have obtained the exact expression of $\Delta v$:
$$\pm \Delta v =  \frac{P\pm \Delta P}{m \sqrt{1+ (P\pm \Delta P)^2/m^2c^2}} -  \frac{P}{m \sqrt{1+ P^2/m^2c^2}},$$
where
$$\Delta P = \frac{\hbar}{2\Delta X}\:.$$
This is a complicated expression but it is easy to see that the final speed cannot exceed $c$ in any cases.
For a fixed value of $P$ and $\Delta X \to 0$, we have
$$v\pm \Delta v = \lim_{\Delta P \to + \infty}\frac{P\pm \Delta P}{m \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}= \pm c\:.\tag{1}$$

Finally, it is not difficult to see that (using the graph of the hyperbolic tangent function)
$$-1 \leq \frac{(P\pm \Delta P)/mc}{ \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}\leq 1\tag{2}\:.$$
We therefore conclude that
$$-c \leq v\pm \Delta v \leq c,$$
where the boundary values are achieved only for $\Delta X \to 0$ according to (1).
Relativity is safe...