A great sphere is totally geodesic

On ${\mathbb{R}}^{r+1}$ the function $q(z)=⟨z,z⟩$ has differential $d{q}_z(w)=2⟨z,w⟩$, which is nonzero on $q^{-1}(1)$. Thus [F2] makes each $S^r$ an embedded boundaryless hypersurface with $T_x{S}^r=x^{\perp }$. Consequently at $x\in S^k$ one has $T_x{S}^k=V\cap x^{\perp }\subseteq T_x{S}^n$, the equatorial inclusion is embedded with the usual induced round metric, and its normal space inside $T_x{S}^n$ is exactly $V^{\perp }$: this subspace lies in $x^{\perp }$, is orthogonal to $V\cap x^{\perp }$, and has the required dimension $n-k$.

For each standard basis vector $a\in \{e_{k+2},\dots ,e_{n+1}\}\subseteq V^{\perp }$, define the smooth tangent field $A(y)=a-⟨a,y⟩y$ on $S^n$. Along $S^k$ one has $⟨a,x⟩=0$, so $A(x)=a$ and this restriction is a normal field to $S^k$ inside $S^n$ by step 1.1. For $X\in T_x{S}^k\subseteq V$, Cartesian differentiation gives $D_{X}A=-⟨a,X⟩x-⟨a,x⟩X=0$. By [F3]–[F4], ${\bar{\mathrm{\nabla }}}_X^{S^n}A=(D_{X}A{)}^{\mathrm{\top }}=0$. Hence the shape operator for $S^k\subseteq S^n$ is $S_{a}X=-({\bar{\mathrm{\nabla }}}_X^{S^n}A{)}^{\mathrm{\top }}=0$.

For tangent vectors $X,Y$, [F5] and step 2.1 give $⟨\mathrm{I}\mathrm{I}(X,Y),a⟩=⟨S_{a}X,Y⟩=0$ for every displayed basis vector $a$ of $V^{\perp }$. By step 1.1 the vector $\mathrm{I}\mathrm{I}(X,Y)$ itself lies in $V^{\perp }$, so it is zero. Thus $\mathrm{I}\mathrm{I}=0$, and [F6] proves that $S^k$ is totally geodesic in $S^n$.

Every admitted sphere is nonempty. For $k=0$, its tangent bundle is zero and the calculation makes $\mathrm{I}\mathrm{I}$ the zero bilinear form. For $k=n$, the displayed normal basis is empty and the normal bundle has rank zero, so $\mathrm{I}\mathrm{I}=0$ automatically. The proof includes $k=1$ and all other intermediate dimensions; the round metrics are positive definite and both manifolds are boundaryless. Only the finite, explicitly displayed normal basis is used. The stated ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through [F2], [F4]–[F6], and the calculation adds no choice. There is no interval, endpoint, or biconditional claim.