A morphism of sheaves is an isomorphism exactly when it is an isomorphism on every stalk

If $\phi$ is an isomorphism with inverse $\psi$, then for every $x$ the induced maps ${\phi }_x$ and ${\psi }_x$ are inverse because taking stalks preserves composition and identities by [F2]. Hence each ${\phi }_x$ is a bijection.

Assume now that every ${\phi }_x$ is a bijection. Fix an open set $U\subseteq X$ and a section $t\in \mathcal{G}(U)$. For each $x\in U$, choose $a_x\in {\mathcal{F}}_x$ with ${\phi }_x(a_x)=t_x$. Pick a representative $s_x\in \mathcal{F}(V_x)$ of $a_x$ on some open neighbourhood $V_x\subseteq U$ of $x$. Then $({\phi }_{V_x}(s_x){)}_x=t_x$, so [F3] lets us shrink $V_x$ and assume ${\phi }_{V_x}(s_x)=t{\mathrm{\mid }}_{V_x}$.

If $s,s^{\mathrm{\prime }}\in \mathcal{F}(U)$ satisfy ${\phi }_U(s)={\phi }_U(s^{\mathrm{\prime }})$, then for every $x\in U$ one has ${\phi }_x(s_x)={\phi }_x(s_x^{\mathrm{\prime }})$ by [F2]. Injectivity of ${\phi }_x$ gives $s_x=s_x^{\mathrm{\prime }}$. By [F3], for each $x$ there exists an open neighbourhood $W_x\subseteq U$ with $s{\mathrm{\mid }}_{W_x}=s^{\mathrm{\prime }}{\mathrm{\mid }}_{W_x}$. The sets $W_x$ cover $U$, so [L2] gives $s=s^{\mathrm{\prime }}$. Thus each ${\phi }_U$ is injective.

Let $x,y\in U$. For any $z\in V_x\cap V_y$, the two sections $s_x{\mathrm{\mid }}_{V_x\cap V_y}$ and $s_y{\mathrm{\mid }}_{V_x\cap V_y}$ have images under $\phi$ whose germs at $z$ both equal $t_z$. Since ${\phi }_z$ is injective, the germs of $s_x$ and $s_y$ at $z$ are equal. By [F3], there is a neighbourhood $W_z\subseteq V_x\cap V_y$ of $z$ on which $s_x{\mathrm{\mid }}_{W_z}=s_y{\mathrm{\mid }}_{W_z}$. The sets $W_z$ cover $V_x\cap V_y$, so [L2] gives $s_x{\mathrm{\mid }}_{V_x\cap V_y}=s_y{\mathrm{\mid }}_{V_x\cap V_y}\mathrm{.}$ Therefore the family $(s_x)$ is compatible on the cover $\{V_x{\}}_{x\in U}$ of $U$.

By [L2], the compatible family of step 2.1 glues to a unique section ${\psi }_U(t)\in \mathcal{F}(U)$ satisfying ${\phi }_U({\psi }_U(t))=t$. Thus ${\phi }_U$ is surjective for every $U$.

Let $V\subseteq U$ and $t\in \mathcal{G}(U)$. Then ${\phi }_V({\psi }_U(t){\mathrm{\mid }}_V)={\phi }_U({\psi }_U(t)){\mathrm{\mid }}_V=t{\mathrm{\mid }}_V={\phi }_V({\psi }_V(t{\mathrm{\mid }}_V))\mathrm{.}$ Step 1.3 gives injectivity of ${\phi }_V$, so ${\psi }_U(t){\mathrm{\mid }}_V={\psi }_V(t{\mathrm{\mid }}_V)$. Hence the maps ${\psi }_U$ commute with restriction and define a morphism of sheaves $\psi :\mathcal{G}\to \mathcal{F}$. By steps 3.1 and 1.3, $\psi$ is a two-sided inverse to $\phi$, so [F1] shows that $\phi$ is an isomorphism. Together with step 1.1 this proves both directions.