Let $(M,g)$ be a complete riemanian manifold with $\text{diam}(M,g) < \infty$. If there are two points $p,q$ s.t. $d(p,q) = \text{diam}(M,g)$ then there are at least two shortest paths between them.
So my idea is to take some geodesic (exists because of completeness) between p and q, lets call it $\gamma: [0,1] \to M$. Then find another geodesic from p to $\gamma(1 + \epsilon)$ which has to exists because of completeness and can not be the same as $\gamma$ since the length has to be $\le \text{diam}(M,g)$.
Then what I want to do is look back at the tangent space $T_pM$ at $p$ and find something like a Cauchy-Sequence of the $v_\epsilon$, where $\exp(tv_\epsilon)$ is the geodesic from $p$ to $\gamma(1+\epsilon)$ and see that the limit of these $v_\epsilon$ gives me another geodesic. But this is where I am stuck. It is not clear that a Cauchy-Sequence like that will exist to me and that the limit does the right thing.
I will tell you how to solve the part where you are stuck, but not the original problem. Your approach is, in fact correct. If $\gamma_\varepsilon$ is the minimizer with endpoint $\gamma(1+\epsilon)$ (starting at that point), parametrized by arclength (!), then the preimage of $\gamma_{\varepsilon}(0)$ under the exponential will converge tothe origin of $T_{\gamma(1)}M$, which corresponds to $\gamma(1)$. Now note that the initial vectors have all length $1$, and that the set of these vectors is compact. This means that a subsequence will converge to some $v\in T_{\gamma(1)}M$ .
If you then look at the geodesic which starts in $\gamma(1)$ in direction $v$ this will be a minimizing (why?) geodesic between the two points you started out with.
Now the tricky part (which I will not solve for you) is to show that the limit $v$ can be chosen in such a way that this limiting geodesic is different from $\gamma$.