Just so I can use this later, let me start by saying that obviously $X$ has to be irreducible for this to be interesting.

Next, I am not entirely sure what a *movable divisor* means. 

I believe that $\mathrm{Mov}(X)$ is usually used to denote the cone of movable **curves**. (See Lazarsfeld's book). The definition of a *movable curve* is that its equivalent (as a $1$-cycle) to the push-forward via $\alpha$ of a complete intersection curve on $X'$, where $\alpha:X'\to X$ is a projective birational morphism.

> As Mark explains in a comment below, his definition of a movable divisor to be one with a base locus of at least codimension two makes sense. Apparently this is used by various authors which makes this one another one of those terrible notations when you won't know what the author really mean until they give you some clues...
In any case, a basepoint-free divisor should be considered movable, so I will give an example for that. 

For a basepoint-free system, first notice that if the Kodaira-dimension of the linear system is at least $2$, then a general member of the very ample linear system whose pull-back is our original linear system will be irreducible and hence so will be a general member of our linear system. So the only chance is with a $1$-Kodaira-dimensional linear system. (By Kodaira-dimension I mean the dimension of the image of $X$).

So, finally, here is an example. Allen's suggestion of two points on a curve is going the right direction, but it doesn't work. If the genus is at least $2$, then it will not move, if it is at most $1$, then its linear system contains a double point. I suppose the next idea is to find an example of a complete linear system on a curve which is basepoint-free, but has no member which is supported at a single point. I would expect that such linear systems exist and perhaps one could construct one with the clever use of étale covers or other tricks. However, I think, one can get an example in a cheaper way: 

Let $Y$ be a smooth projective curve (say over an algebraically closed field) of some high genus and $L$ a very very ample linear system. No matter what, there will be only finitely many (zero is finite!) points, say $\{P_1,\dots,P_m\}\subset Y$ such that if $L$ contains a member supported on a single point, then that point is one of the $P_i$'s. Next take an arbitrary projective surjective morphism $f:Z\to Y$ with connected fibers and pick points $\{Q_1,\dots,Q_m\}\subset Z$ such that $f(Q_i)=P_i$. Finally, let $X$ be the blow up of $Z$ along $\{Q_1,\dots,Q_m\}$ and consider the linear system $\mathfrak d$ which is the pull-back of $L$ on $X$. Now, any member of $L$ which is not supported at a single point pulls-back to a reducible divisor. So do those supported at the $P_i$'s because of the blow-up. Finally, since the fibers of the morphism $X\to Y$ are connected, all the members of $\mathfrak d$ are actual pull-backs (just look at $h^0$ of the corresponding line bundles), so we have accounted for all the members.