= Solution
By part (b), $M=S^n/\Gamma$, where the group of <deck transformations> acts freely by round <isometries>. Every round <isometry> is the restriction of an <orthogonal transformation> of $\mathbb R^{n+1}$: it preserves the inner product $\langle x,y\rangle=\cos d(x,y)$, so its images of an orthonormal basis determine its linear extension.
Suppose $n$ is even. Every $A\in SO(n+1)$ has eigenvalue one. Indeed, the nonreal <eigenvalues> come in conjugate pairs with product one, while the remaining real eigenvalues are $\pm1$. Their number is odd and their product is one, forcing at least one $+1$. Such a transformation fixes a unit vector. Freeness of the <deck transformation> action consequently implies that the <determinant> map $\Gamma\to\{\pm1\}$ has trivial kernel. Thus $|\Gamma|\leq2$.
If $\Gamma$ is nontrivial with generator $A$, then $A^2=I$. Its eigenvalues are $\pm1$, and freeness excludes the eigenvalue $+1$. Hence $A=-I$, the antipodal <involution>. We obtain
$$
\boxed{M\cong S^n\quad\text{or}\quad M\cong\mathbb{RP}^n.}
$$
This proves the even-dimensional part of the <classification of complete positive constant-curvature manifolds>.
\b[The conclusion fails in odd dimensions.] For $n=2m-1\geq3$, identify $S^{2m-1}$ with the unit sphere in $\mathbb C^m$. Scalar multiplication by $\zeta=e^{2\pi i/r}$, with $r\geq3$, generates a freely acting <cyclic group> of <isometries>: $\zeta^jz=z$ for nonzero $z$ forces $\zeta^j=1$. The quotient is complete with constant <sectional curvature> one and has <fundamental group> $C_r$, so it is neither the simply connected <sphere> nor <Real projective space> with fundamental group $C_2$. In particular $S^3/C_3$ is a <lens space> counterexample.
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