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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.

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 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.

If we take that as a definition, then a movable divisor could only exist on a surface and it would be the push-forward of a very ample class and hence a general member of its linear system would be irreducible. So, that's out.

For divisors in general I am not too enthusiastic of saying that something is movable if its base locus is small. That would mean that the pull-back of a movable divisor via a birational morphism (say blowing up the base locus) is no longer movable. I wonder if such a notion has any reasonable usefulness. (Seriously, I wonder, so if you know, let me know!).

I could imagine two reasonable directions for a definition for a movable divisor:

1. Something about the linear system not consisting only of that one divisor. Say that the dimension of the linear system is positive. This, of course, allows fixed components.
2. Basepoint-freeness. This is actually the context I have seen this used. Say someone takes a linear system and in order to separate the base points from the linear system, blows up the base locus. The pull-back of the original linear system now has a fixed part and (what is usually called) a movable part. This movable part is basepoint-free and in fact, this is the one giving the morphism which is the resolution of indeterminacies of the rational map determined by the original linear system.

So, let's see if we can find examples using these definitions.

The first one will obviously work, but it's kind of boring: Take any linear system with a fixed part and some non-trivial moving part. Then each member will be reducible. I suspect this is not what the OP was hoping for.

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.

1. Something about the linear system not consisting only from that one divisor. Say that the dimension of the linear system is positive. This, of course, allows fixed components.
2. Basepoint-freeness. This is actually the context I have seen this used. Say someone takes a linear system and in order to separate the base points from the linear system, blows up the base locus. The pull-back of the original linear system now has a fixed part and (what is usually called) a movable part. This movable part is basepoint-free and in fact, this is the one giving the morphism which is the resolution of indeterminacies of the rational map determined by the original linear system.

1. Something about the linear system not consisting only of that one divisor. Say that the dimension of the linear system is positive. This, of course, allows fixed components.
2. Basepoint-freeness. This is actually the context I have seen this used. Say someone takes a linear system and in order to separate the base points from the linear system, blows up the base locus. The pull-back of the original linear system now has a fixed part and (what is usually called) a movable part. This movable part is basepoint-free and in fact, this is the one giving the morphism which is the resolution of indeterminacies of the rational map determined by the original linear system.