 Solution

The kernel of the restriction mapconsists exactly of sections whose germs vanish outside $Z$, namely ${\Gamma }_Z(X,\mathcal{F})$. This proves exactness. If $\mathcal{F}$ is a flasque sheaf, the restriction map is surjective by definition.

Now let $0\to {\mathcal{F}}_1\to {\mathcal{F}}_2\to {\mathcal{F}}_3\to 0$ be exact. The global section functor is left exact, so a section of ${\Gamma }_Z(X,{\mathcal{F}}_2)$ mapping to zero lifts uniquely to a global section of ${\mathcal{F}}_1$. Its germs outside $Z$ vanish because ${\mathcal{F}}_1\to {\mathcal{F}}_2$ is injective at each stalk of a sheaf. It therefore lies in ${\Gamma }_Z(X,{\mathcal{F}}_1)$, proving exactness of the supported-section sequence.

Suppose in addition that ${\mathcal{F}}_1$ is flasque and take $s_3\in {\Gamma }_Z(X,{\mathcal{F}}_3)$. Surjectivity on global sections gives a lift $s_2\in \Gamma (X,{\mathcal{F}}_2)$. On $X\setminus Z$, its restriction comes from some $t_1\in \Gamma (X\setminus Z,{\mathcal{F}}_1)$ by left exactness. Extend $t_1$ to ${\tilde{t}}_1\in \Gamma (X,{\mathcal{F}}_1)$ using flasqueness. Then $s_2-{\tilde{t}}_1$ maps to $s_3$ and vanishes outside $Z$, proving surjectivity on the right.