Let $X$ be an irreducible smooth curve, and $\underline{k(X)}$ the constant sheaf on $X$ with the function field $k(X)$ as fibers.
Reading from Serre's Algebraic groups and class fields I met the claim ($H$ denotes sheaf cohomology) $$ H^1(X, \underline{k(X)}) = 0 $$ motivated by the sentence: "the nerve of every open cover of $X$ is a simplex".
I don't know what the nerve of an open cover is, but anyway the above claim seems quite intuitive: an element of $H^1$ is a section of $\underline{k(X)}$ defined only on the intersection of pairs of opens of a cover of $X$. Such a local section, in our case, always comes from the restriction of a globally defined section. After all the sheaf is constant, and a global section consists of a collection of an arbitrary regular function for every point of $X$, with any kind of jumps allowed.
Is my reasoning correct? Could you please help me to formalize it?
Result
If $X$ is an irreducible topological space and $\underline A$ the constant sheaf associated to the abelian group $A$, then we have for the first Čech cohomology group $\check H^1(X,\underline A)=0$ .
Proof
It suffices to prove that, for an arbitrary covering $\mathcal U =(U_i)_{i\in I}$ we have $ \check H^1(\mathcal U,\underline A)=0$.
Given a cocycle $(g_{ij})\in Z^1(\mathcal U,\underline A)$, fix some $i_0\in I$ and define the $0$-cochain $(h_i)\in C^0(\mathcal U,\underline A)$ by $h_i=g_{i_0i}$. The cocycle condition $g_{i_0j}=g_{i_0i}+g_{ij}$ translates into $h_j=h_i+g_{ij}$ or $g_{ij}=h_j-h_i$.
This proves that the cocycle $(g_{ij})$ is the coboundary of the cochain $(h_i)$, and thus that the class of that cocycle is zero in $ \check H^1(\mathcal U,\underline A)$.
Since $(g_{ij})$ was an arbitrary cocycle, we have $ \check H^1(\mathcal U,\underline A)=0$, as promised.
Check your understanding
Where on earth did I use that our sheaf is constant and that $X$ is irreducible ?!
Edit: Answer
When I defined $h_i=g_{i_0i}$.
Since $g_{i_0i}\in \Gamma(U_{i_0i},\underline A)$ and $h_i\in \Gamma(U_i,\underline A)$, the definition $h_i=g_{i_0i}$ only makes sense because $\Gamma(U_i,\underline A)=\Gamma(U_{i_0i},\underline A)=A$.
See Remark 1) below where I emphasize that $\Gamma(U,\underline A)=A$ for all open non-empty $U$'s.
Remarks
1) The irreducibility of $X$ implies that the constant presheaf associated to $A$ is already a sheaf i.e. that the constant sheaf $\underline A$ is given by $\Gamma(U,\underline A)=A$ for all open non-empty subsets $U\subset X$.
(For reducible topological spaces $\Gamma(U,\underline A)$ consists of locally constant maps $U \to A$)
This implies that $\underline A$ is flabby (=flasque).
2) For arbitrary flabby sheaves $\mathcal F$ on arbitrary topological spaces $Y$ both Čech cohomology groups and genuine cohomology groups (as introduced by Grothedieck through derived functors) are zero in all positive degrees : $\check {H}^n(Y,\mathcal F)=H^n(Y,\mathcal F)=0$ for $n\gt 0$.
3) And to finish on a pleasant note: Čech cohomology groups and genuine cohomology groups always coincide in degree $1$ for all sheaves $\mathcal G$ of abelian groups, flabby or not: $\check {H}^1(Y,\mathcal G)=H^1(Y,\mathcal G)$.