Constant Presheaf not necessarily a sheaf. Proof? Let $\mathcal{X}$ be a topological space and $\mathcal{F}$ the constant presheaf, that assigns to each open set $\mathcal{U}$ of $\mathcal{X}$ the set $A$. The restriction map is the identity $A \rightarrow A$. I want to show that this presheaf is not necessarily a sheaf and I am trying to understand the argument in this wikipedia article: http://en.wikipedia.org/wiki/Constant_sheaf.
In particular, let $\emptyset = \cup_i \mathcal{U}_i, \, \mathcal{U}_i=\emptyset$, be a covering of the empty set. Take two sections over $\emptyset$, i.e. take $s,s' \in \mathcal{F}(\emptyset)=A$. The wikipedia article says that the restrictions of $s,s'$ to any $\mathcal{U}_i$ must be equal. I don't understand this. For example, since $\mathcal{U}_i \subset \emptyset$, then we have a restriction map $\mathcal{F}(\emptyset) \rightarrow \mathcal{F}(\mathcal{U}_i)$ which sends $s$ to $s$ and $s'$ to $s'$. So why $s=s'$?
Something else, possibly related to what I am missing: since the elements of $A$ are abstract, we can not really say anything about $s|_{\mathcal{U}_i}$, right?

Thanks.
The argument in the Wikipedia article states that you should take the empty covering of the empty set, i.e. your indexing set (where $i$ comes from) is the empty set. The empty union is the empty set, so it is a covering.

Then you pick any $s \neq s^\prime \in A$ and consider them to be sections over the empty set ($A$ needs to have at least two elements). The condition that they agree after restricting is an empty one (because there is no set to restrict to), so by the local identity axiom, you would need $s=s^\prime$, a contradiction.
I think you are taken the wrong definition of the constant presheaf. Since the presheaf $\mathcal{F}$ is a functor on the category of open sets of $X$, the constant presheaf means that $\mathcal{F}$ is a constant function, i.e., $\mathcal{F}(U)= \mathcal{F}(V)$ for all $U,V$ open in $X$.