Locally constant sheaf but not constant

If $X$ is locally connected, locally constant sheaves are (up to isomorphism) exactly the sheaves of sections of covering spaces $\pi:Y\to X$.
Such a locally constant sheaf is a constant sheaf if and only the covering $\pi$ is trivial.
So any non trivial covering will give you a non-constant but locally constant sheaf.
The simplest example is the sheaf of sections of the two sheeted non trivial covering $\mathbb C^*\to \mathbb C^*:z\mapsto z^2$ or its restriction to the unit circle $S^1\to S^1: e^{i\theta}\mapsto e^{2i\theta}$

Suppose that $X$ is connected (and is reasonable enough that the usual theory of covering spaces applies), and choose a base-point $x \in X$. Then giving a locally constant sheaf (of abelian groups, say) on $X$ is the same as giving an abelian group $M$ with an action of $\pi_1(X,x)$.

The group $M$ arises as the stalk of the locally constant sheaf at $x$, and the action of $\pi_1(X,x)$ is obtained by choosing an element $m \in M$, think of $m$ as a section in a n.h. of $x$, and then move the section around loops based at $x$ by using the locally constant structure. (This is called the monodromy action of $\pi_1(X,x)$ on the stalk $M$.)

Note: this is a rephrasing of Georges Elencwajg's answer, using the language of $\pi_1$ rather than the language of covering spaces (although the latter is implicit in my answer).

There even exist examples where $X$ is connected. For instance, consider the covering map $\mathbf R\to S^1$ given by the exponential. The sheaf of continuous sections of this map is a locally constant sheaf of sets on $S^1$, but it is not constant because it has no global section.