$X$ being locally closed is not equivalent to every extension by $0$ of a sheaf $F$ being unique? In Tennison's book "Sheaf Theory", the author presents a proof that there is a unique extension by $0$ for a sheaf $F$ over $X$ iff $X \subset Y$ is locally closed . However, apparently in the proof of the "if" part, he uses that some neighborhood of a given point is connected, therefore the proof is wrong. I would like to know if this fact is indeed correct.
I think that this theorem is not true, because picking $Y = \mathbb{R}^2$ and $X$ equals an open disk, then the sheaf $\mathbb{Z}_X$ can be extended to the whole plane by picking the sheaffication of $F (U) =\mathbb{Z}$ if $U \subset X$ and $F = 0$ otherwise, then each stalk lying out of the disk is zero. Furthermore any other sheaf  extending by $0$ $\mathbb{Z}_X$ (apparently) will be isomorphic to this one. But $D = X$ is open , hence not locally closed!!!! Am I missing something?
Thanks in advance.
 A: In fact the theorem is true, because the presheaf and the sheaf have the same stalks, hence  by comparing some with a constant section such that they have the same value in the stalk one gets the neighborhood.
For the case of the sheaf $\mathbb{Z}_X$ where $X = \mathbb{R}^2$, the inclined disk (as said in the commentaries) is, in fact, not a section since this set in the étale space is not open (otherwise intersecting this with an horizontal open disk would be an open set, however it will be a line homeomorphic to a line in $\mathbb{R}^2$, an absurd).
Why does the stalk APPARENTLY fail in converging to the integers?
My confusion about the connectedness was given by the fact that the value of the sheaf in two disjoint disks is $\mathbb{Z}\oplus \mathbb{Z}$. Therefore, when the space is tottaly disconnected, this does happens with every element of the basis (of open sets).
For instance, let $X = \mathbb{R}_S$ be the Sorgenfrey line and consider the constant sheaf over it. By letting $I_n = [0, 1/n[$, one can see that, at each $I_n$, the section can attains just a constant value, therefore $\mathbb{Z}_X (I_1)$ is at least $\mathbb{Z}^{\omega}$. Looking the way mentioned above, it looks like that, at the stalk (in $0$), the sheaf will not converges to $\mathbb{Z}$, since the diagram of the colimit will looks like a lot of arbitrary large products of the integers (at least as subgroups).
Why does it, in fact, hold?
However, fixing a section one can see that it must attains a fixed value in a neughborhood of the $0$ (because the constant function is a section), and then letting it "flows" throught the diagram, it´s easily seen that it will be an integer. Therefore , in fact, $(\mathbb{Z}_X)_0 = \mathbb{Z}$.
