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I would guess that the compact spaces you are looking for are extremely rare.

For example the extremely simple contractible space $I=[0,1]$ is not suitable:
Consider the inclusion $j\colon U=(0,1)\hookrightarrow I $ and take on $I$ the sheaf$j_!(\mathbb Z_U)$, the constant sheaf $\mathbb Z_U$ on $U$ extended to $I$ by zero.
Claim: $H^1(I,F)\cong\mathbb Z$
Proof of claim:
Consider the open embedding $i:F=I\setminus U\hookrightarrow X$ and the short exact sequence of sheaves on $I$ (see Hartshorne's Algebraic Geometry, Exercise I.19, page 68) : $$0\to j_!\mathbb Z_U\to \mathbb Z_I\to i_*\mathbb Z_F\to 0$$ Taking the corresponding long exact sequence in cohomology we get the fragment $$0\to \Gamma(I,j_!\mathbb Z_U) \to \Gamma(I,\mathbb Z_I)\to \Gamma(I,i_*\mathbb Z_F)\to H^1(I,j_!\mathbb Z_U) \to H^1(I,\mathbb Z_I) $$ Since $\Gamma(I,j_!\mathbb Z_U)=0$ and $H^1(I,\mathbb Z_I)=H^1_{\operatorname {singular}}(I,\mathbb Z)=0$ the above fragment becomes $$0\to 0\to \mathbb Z \to \mathbb Z^2\to H^1(I,j_!\mathbb Z_U) \to 0 $$ so that $H^1(I,j_!\mathbb Z_U)\cong \mathbb Z\neq 0$

There is a very similar statement in scheme theory saying that $H^1(\mathbb A^1_k,j_!(\mathbb Z_U))=\mathbb Z$, where now $U$ is the complement of two rational points on the affine line $\mathbb A^1_k$: see Hartshorne's Algebraic Geometry, Exercise III 2.1

Of course on afffine schemes, quasi-coherent sheaves have zero cohomology in positive degree, but that is not a purely topological statement and as shown in the example 2) above does not apply to arbitrary sheaves of abelian groups: even theorems by Serre necessitate some hypotheses!