= Solution
Let $t_0$ be the finite cut time and put $q=\gamma(t_0)$. Continuity of <Riemannian distance> shows that the segment to $q$ still minimizes and $d(p,q)=t_0$. Choose $t_j\downarrow t_0$ with $t_j>t_0$. By the <Hopf-Rinow theorem>, there is a minimizing <geodesic> from $p$ to $q_j=\gamma(t_j)$ with length $\ell_j=d(p,q_j)<t_j$ and initial unit vector $v_j$. Continuity gives $\ell_j\to t_0$.
The unit tangent sphere is compact, so after a subsequence $v_j\to v$. Smooth dependence of the <Riemannian exponential map> gives $\exp_p(t_0v)=q$. The <geodesic> with initial vector $v$ has length $t_0=d(p,q)$ up to $q$, hence minimizes. If $v\ne\dot\gamma(0)$, this is the required second minimizing <geodesic>.
Suppose instead $v=\dot\gamma(0)$ and $q$ is not a <conjugate point>. Then the <differential> of $\exp_p$ at $t_0\dot\gamma(0)$ is nonsingular. The <inverse function theorem> makes $\exp_p$ injective on a neighborhood of this vector. For large $j$, both $\ell_jv_j$ and $t_j\dot\gamma(0)$ belong to that neighborhood and exponentiate to $q_j$. They must be equal, implying $\ell_j=t_j$, a contradiction. Thus either a distinct minimizing <geodesic> exists or $q$ is conjugate. In the latter case it is the first <conjugate point>, since part (a) excludes any earlier one on a minimizing segment. We have proved the \b[cut-point dichotomy for complete Riemannian manifolds].
Here the relevant concept is a <Riemannian cut point>, not the topological <cut point> obtained by disconnecting a space through removal of a point.
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