An example that Čech cohomology is not equal to derived cohomology with on an affine scheme with Zariski topology. There is an example in 3.1.10 $\mathbb{A}^1$-homotopy theory of schemes demonstrating that the Čech cohomology on an affine scheme with Zariski topology can be different from the derived cohomology.
Let $x_1,x_2 \in \mathbb{A}^2$ be two distinct closed points and let $S$ be the semilocal ring at $x_1$ and $x_2$. Let $C_1,C_2 \subset \mathbb{A}^2$ be two irreducible curves intersecting transversely at $x_1,x_2$. Let $U= \mathrm{Spec}\, S\backslash(C_1\cup C_2), V =  \mathrm{Spec}\, S \backslash\{x_1,x_2\}$.  Let $j$ be the inclusion $U \rightarrow \mathrm{Spec}\, S$.
We have the decomposition
$$V= (V\backslash(V\cap C_1))\cup (V\backslash(V\cap C_2))$$
whose Mayer-Vietoris sequence give
$$0=H^0(V\backslash(V\cap C_1),j_!\mathbb{Z})\oplus H^0(V\backslash(V\cap C_2),j_!\mathbb{Z})\rightarrow H^0(U,j_!\mathbb{Z}) = H^0(U,\mathbb{Z})=\mathbb{Z}\rightarrow H^1(V,j_!\mathbb{Z}).$$
Therefore, $H^1(V,j_!\mathbb{Z})\neq 0$. The decomposition
$$\mathrm{Spec}\, S = (\mathrm{Spec}\, S\backslash x_1 )\cup (\mathrm{Spec}\, S\backslash x_2 )$$
gives a Mayer-Vietoris sequence
$$H^1(\mathrm{Spec}\, S\backslash x_1,j_!\mathbb{Z})\oplus H^1(\mathrm{Spec}\, S\backslash x_2,j_!\mathbb{Z})\rightarrow H^1(V,j_!\mathbb{Z}) \rightarrow H^2(S,j_!\mathbb{Z}).$$
Then he conclude that $H^2(S,j_!\mathbb{Z})\neq 0$. But how can he says that the image of the nonzero element in $H^1(V,j_!\mathbb{Z})\neq 0$ is not zero in $H^2(S,j_!\mathbb{Z})$? (He says that this is because $C_1$ and $C_2$ are irreducible. But I don't know how this can leads to the conclusion.)
 A: Write $S$ for the spectrum of the semi-local ring our considering, $S_i := S \setminus \{x_i\}$, and $D_i := S_i \cap (C_1\cup C_2) = S_i \setminus U$.
Then, I think $H^1(S_i,j_!\mathbb{Z}) = 0$.
So the morphism $H^1(V,j_!\mathbb{Z}) \to H^2(S,j_!\mathbb{Z})$ is injective.
I would suggest you to consider the following exact sequence on $S_i$:
$$
0 \to j_! \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}|_{D_i} \to 0.
$$
Since $\mathbb{Z}$ is a constant sheaf, and any open subset of $S_i$ is irreducible (in particular, connected), $\mathbb{Z}$ is flasque.
Hence $H^l(S_i,\mathbb{Z}) = 0$ for any $l \geq 1$, and $H^0(S_i,\mathbb{Z}) = \mathbb{Z}$.
Moreover, since $D_i$ is connected, $H^0(S_i,\mathbb{Z}|_{D_i}) = H^0(D_i,\mathbb{Z}) = \mathbb{Z}$.
These imply that $H^1(S_i,j_!\mathbb{Z}) = 0$ (note that the morphism $\mathbb{Z} = H^0(S_i,\mathbb{Z}) \to H^0(S_i,\mathbb{Z}|_{D_i}) = \mathbb{Z}$ is induced by $\mathbb{Z} \to \mathbb{Z}|_{D_i}$, hence this is an isomorphism):
$$
[H^0(S_i,\mathbb{Z}) = \mathbb{Z}] \xrightarrow{1\mapsto 1} [H^0(S_i,\mathbb{Z}|_{D_i}) = \mathbb{Z}] \to H^1(S_i,j_!\mathbb{Z}) \to [H^1(S_i,\mathbb{Z}) = 0].
$$
