The quotient of the constant sheaf by the structure sheaf is flasque. Let $X$ be an integral scheme of finite type over $k$ with $\dim(X)=1$. Let $\mathcal{K}_{X}$ be the constant sheaf on $X$ with value $K(X)$, where $K(X)$ is the function field of $X$. 
I want to show that the quotient sheaf $\mathcal{G}:=\mathcal{K}_{X}/\mathcal{O}_{X}$ is flasque. 
Since $X$ is integral (and thus irreducible) we clearly see that the constant sheaf itself $\mathcal{K}_{X}$ is flasque. (don't know if this fact will be useful).
We thus want to show that for every open $V\subset U\subset X$ the restriction map $r:\mathcal{G}(U)\rightarrow\mathcal{G}(V)$ is surjective. 
Note that the quotient sheaf is defined as the sheafification of the quotient presheaf mapping an open $U\mapsto \mathcal{K}_{X}(U)/\mathcal{O}_{X}(U)$.
Added Approach: What I tried to show is that $\mathcal{O}_{X}$ is flasque, since then I could use the exact sequence $0\to\mathcal{O}_{X}\to\mathcal{K}_{X}\to\mathcal{K}_{X}/\mathcal{O}_{X}\to 0$ and the fact that $\mathcal{K}_{X}$ is flasque to conclude that $\mathcal{K}_{X}/\mathcal{O}_{X}$ is also flasque. (see comments for a counter example). 
Any help would be appreciated!
 A: Taking sections on any open subset $U\subset X$ and using the fact that flasque sheaves have no higher cohomology we get a long exact sequence $$0\to \mathcal{O}_X(U)\to \mathcal{K}_X(U)\to (\mathcal{K}_X/\mathcal{O}_X)(U) \to H^1(\mathcal{O}_X,U) \to 0.$$
Using the restriction maps, we have the following diagram for each pair $V\subset U\subset X$ (it suffices to consider $V\subsetneq U$ since the restriction map $res_{U,U}=id$ is always surjective):
$$
    \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
    \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
    \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
    \begin{array}{c}
0 & \ra{} & \mathcal{O}_X(U) & \ra{} & \mathcal{K}_X(U) & \ra{} & (\mathcal{K}_X/\mathcal{O}_X)(U) & \ra{} & H^1(\mathcal{O}_X,U) & \ra{} & 0 \\
\da{} & & \da{res_{U,V}} & & \da{res_{U,V}} & & \da{res_{U,V}} & & \da{res^1_{U,V}} & & \da{} \\
0 & \ras{} & \mathcal{O}_X(V) & \ras{} & \mathcal{K}_X(V) & \ras{} & (\mathcal{K}_X/\mathcal{O}_X)(V) & \ras{} & H^1(\mathcal{O}_X,V) & \ras{} & 0 \\
\end{array}
$$
and we would love to use this to conclude that the restriction map on $\mathcal{K}_X/\mathcal{O}_X$ is surjective using one of the weaker versions of the Five lemma. But we need more information before we can apply that lemma. We have that the restriction map $res_{U,V}:\mathcal{K}_X(U)\to\mathcal{K}_X(V)$ is surjective by definition of a flasque sheaf, and the final right-hand map $0\to 0$ is injective, but we need to show that our map $res_{U,V}^1:H^1(\mathcal{O}_X,U)\to H^1(\mathcal{O}_X,V)$ is surjective in order to use the lemma and be done.
This is where the assumptions that $X$ is an integral curve comes in. In an integral curve, every proper open subset $U\subsetneq X$ is affine, so $H^1(\mathcal{O}_X,V)=0$. So $res_{U,V}^1:H^1(\mathcal{O}_X,U)\to H^1(\mathcal{O}_X,V)=0$ is the zero map to the zero module, which is surjective. So we're done.
