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

> **EDIT**: In an earlier version of this answer I ruminated on what a *movable divisor* might mean as it seems to conflict with the definition of a *movable curve*. Following Mark's advice in the comments (thanks!) I unearthed a definition in a [paper of Kawamata][1] which agrees with what Mark is suggesting in another comment. It still seems an unfortunate overlap between the two notions, but probably we can't help it now. 
In any case, a basepoint-free divisor should be considered movable by any definition (and it satisfies the one by Mark et al), so I will give an example of a basepoint-free linear system with the required property.

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. 


  [1]: http://www.jstor.org/stable/1971417