Does the Čech cohomology always give rise to a long exact sequences given a short exact sequence of sheaves?
Clearly that cannot occur for sheaves on a paracomact (perhaps also Hausdorff, I'm not sure about that) space, but the argument one uses there relies on the topological assumptions in a crucial way and cannot be generalized to an arbitrary space, not in a straight-forward manner anyway.
Thank you for shedding any light on this issue!
It should be no. Sorry, I don't have the time to write anything detailed, so I'll just leave assembly instructions. Take a look at my answer here: equivalence of Grothendieck-style versus Cech-style sheaf cohomology describing an example due to Grothendieck where Čech differs from derived functor cohomology. If you write out what you get from the sequence $$0\to K\to \mathbb{Z}_X\to \mathbb{Z}_Y\to 0$$ described there, the Čech sequence should fail to be exact at $\check{H}^2(K)$.
Here are some details. $X$ is the affine plane over a field with the Zariski topology. $Y\subset X$ is the union of two irreducible curves $Y_i$ meeting at two points. If you are not familiar with algebraic geometry, the key point is that the sheaves $\mathbb{Z}_X$, $\mathbb{Z}_{Y_i}$ are flasque hence acylic. From the long exact sequence for sheaf cohomology (= derived functors of global sections), we have an exact sequence $$ H^1(X,\mathbb{Z}_X)=0\to H^1(Y,\mathbb{Z}_Y)\to H^2(X, K)\to H^2(X,\mathbb{Z}_X)=0$$ Using Mayer-Vietor, the second, and therefore third group, is just $\mathbb{Z}$. Now consider the sequence (I'm not saying exact sequence), $$ \check{H}^1(X,\mathbb{Z}_X)\to \check{H}^1(Y,\mathbb{Z}_Y)\to \check{H}^2(X, K)\to $$ It is known that $\check{H}^1$ coincides with $H^1$, but that $\check{H}^2(K)=0$ [cf Grothendieck "Sur quelques points...", sect 3.8]. So exactness of the second sequence fails.
