This is a problem 1.13 of Tennison Sheaf Theory.
"Construct a topological space $X$ and presheaf $F$ of abelian groups on $X$ with the following properties.
(a) for any open $U\subset X$, $F(U)\neq\{0\}$
(b) for all $x\in X$, the stalk $F_x=\{0\}$
(If you cannot, prove that it is impossible.)"
Since $F$ is contravariant functor from category of topological space $X$'s open set to abelian group category, I would like $F$ acting on morphism for any distinct two open sets $V\subset U$, $F(U)\to F(V)$ as trivial map and $F(U)\to F(U)$ is just standard $1_{F(U)}$ map. Then the stalk at any point of $X$ is trivial. This is a presheaf.
However, pictorially, I draw picture of presheaf over $X$ as some kind of vector bundle over $X$ and I suspect that there will be non-trivial stalk somewhere. I should not draw presheaf as vector bundle which is totally different from presheaf.
Which conclusion is correct?
Your example is correct (assuming no element $x\in X$ has a smallest open neighborhood $U$, since then the stalk at $x$ would be $F(U)$ which is nontrivial). To the extent that this example violates your intuition, that means your intuition is incorrect. A presheaf is just any functor on the category of open subsets of $X$, and this is a very flexible notion. It doesn't have to look anything remotely like the "sections" of some kind of "bundle". That intuition is only really appropriate for sheaves.