Pin and full-even Spin are topological double covers

The maps are continuous because the twisted action uses only finite-dimensional algebra multiplication, the grading involution, and inversion on the open set of invertible elements. By [F2], shrink a neighborhood $U$ of the identity in $SO(V,B)$ until the section is defined on $U$. For every $A\in U$, [F1] says the fiber is exactly $\{s(A),-s(A)\}$.

Since the finite-dimensional Clifford algebra is Hausdorff and $1\mathrm{\ne }-1$, choose an open neighborhood $W$ of $1$ disjoint from $-W$. Since $s(I)=1$, shrink $U$ so $s(U)\subset W$ and $-s(U)\subset -W$. Then ${\rho }^{-1}(U)\cap W=s(U)$ and ${\rho }^{-1}(U)\cap (-W)=-s(U)$ are disjoint open subsets of ${\rho }^{-1}(U)$; each maps homeomorphically to $U$, with inverse $s$ or $-s$. Thus the map on full-even Spin is a two-sheeted covering near the identity.

If $n=0$, then [F5] gives $\mathrm{C}\mathrm{l}(V,Q)=\mathbb{R}$, all its elements are even, and the definition gives $\mathrm{Pin}(0)={\mathrm{Spin}}_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}(0)=\{1,-1\}$. The target groups $O(0)$ and $SO(0)$ are both the singleton $\{I_0\}$, and both action maps are the two-to-one map from the discrete two-point group to a point. Thus both maps are covering maps in dimension zero. Now suppose $n>0$. Choose an open neighborhood $U_O$ of the identity in $O(V,B)$, contained in both $SO(V,B)$ and the neighborhood $U$ of step 2.1; this is possible because $SO(V,B)$ is open in $O(V,B)$, as determinant is continuous with values in $\{1,-1\}$. If $x\in \mathrm{Pin}$ and $x=u_1\cdots u_k$ is a product of normalized vectors, then [F4] gives $\det {\rho }_x=(-1{)}^k$. Thus ${\rho }_x\in U_O$ implies $k$ is even, and [F3] gives $x\in {\mathrm{Spin}}_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$. Conversely full-even Spin is contained in Pin. Therefore the Pin preimage of $U_O$ is exactly the full-even Spin preimage restricted to $U_O$, so step 2.1 supplies its two sheets. For any point of either target group choose one lift using [F1]; left multiplication by that lift and by its image transports the identity neighborhood and its two sheets to a neighborhood of that point. Hence both maps are covering maps everywhere. Their kernels are central by [F1], so both are central topological double covers.