Why does a path homotopy between distinct geodesics must pass through a long curve?

I'm working on problem from Do Carmo's "Riemannian Geometry" (Ex. 1, chapter 10). The problem, called Klingenberg's Lemma, puts a lower bound on the length of some curve in a homotopy, provided the homotopy is between distinct geodesics in a complete manifold with bounded sectional curvature.

This question already deals with some aspects of the same problem, so I'll quote some of it. My thanks to @Colescu for the monumental typing effort. Here's the problem, plus the provided hints:

(Klingenberg's Lemma). Let M be a complete Riemannian manifold with sectional curvature $$K, where $$K_0$$ is a positive constant.
Let $$p,q \in M$$ and let $$\gamma_0,\gamma_1$$ be two distinct geodesics joining p to q with $$\ell[\gamma_0]\leq \ell[\gamma_1]$$.
Let $$\alpha_t$$ be a homotopy from $$\gamma_0$$ to $$\gamma_1$$.
Prove that there exists $$t_0\in[0,1]$$ such that $$\ell[\gamma_0]+\ell[\alpha_{t_0}]\geq \frac{2\pi}{\sqrt{K}}$$
Hint: Assume $$\ell(\gamma_0)<\pi/\sqrt{K_0}$$ (otherwise, we have nothing to prove). From Rauch's Theorem, $$\exp_p:TpM\to M$$ has no critical point in the open ball $$B$$ of radius $$\pi/\sqrt{K_0}$$, centered at $$p$$. For $$t$$ small, it is possible to lift the curve at to the tangent space $$T_pM$$, i.e., there exists a curve $$\widetilde{\alpha}_t$$ in $$T_pM$$, joining $$\exp_p^{-1}(0)=0$$ to $$\exp_p^{-1}(q)=\widetilde{q}$$, such that $$\exp_p\circ\widetilde{\alpha}_t=\alpha_t$$. It is clear that it is not possible to do the same for every $$t\in[0,1]$$, since $$\gamma_1$$ cannot be lifted keeping the endpoints fixed.

We conclude that for all $$\varepsilon>0$$ there exists a $$t(\varepsilon)$$ such that $$\alpha_{t(\varepsilon)}$$ can be lifted to $$\tilde{\alpha}_{t(\varepsilon)}$$ and $$\tilde{\alpha}_{t(\varepsilon)}$$ contains points with distance $$<\varepsilon$$ from the boundary $$\partial B$$ of $$B$$. In the contrary case, for some $$\varepsilon>0$$, all lifts $$\tilde{\alpha}_t$$ are at the distance $$\geq\varepsilon$$ from $$\partial B$$; the set of $$t$$'s for which it is possible to lift $$\alpha_t$$ will then be open and closed and $$\alpha_1$$ could be lifted, which is a contradiction.

So far, I've managed to cope - some adjustments to standard proofs of the path-lifting and homotopy lifting properties of cover maps generalize for a non-singular $$\exp_p$$, under some constraints.

The next part of the hint still bugs me:

Therefore, for all $$\varepsilon>0$$, we have $$\ell(\gamma_0)+\ell(\alpha_{t(\varepsilon)})\geq\frac{2\pi}{\sqrt{K_0}}-\varepsilon.$$ Now choose a sequence $$\{\varepsilon_n\}\to0$$, and consider a convergent subsequence of $$\{t(\varepsilon_n)\}\to t_0$$. Then there exists a curve $$\alpha_{t_0}$$ with $$\ell(\gamma_0)+\ell(\alpha_{t_0})\geq\frac{2\pi}{\sqrt{K_0}}.$$

I can see why, since $$\tilde{\alpha}_{t(\varepsilon)}$$ contains points with distance $$<\varepsilon$$ from the boundary $$\partial B$$ of $$B$$, we have: $$\ell[\alpha_t]\geq \frac{\pi}{\sqrt{K_0}}-\varepsilon$$ But since we assume $$\ell(\gamma_0)<\pi/\sqrt{K_0}$$, I can't derive the inequality: $$\ell(\gamma_0)+\ell(\alpha_{t_0})\geq\frac{2\pi}{\sqrt{K_0}}-\varepsilon$$

I've tried to sharpen the bound on $$\ell[\alpha_t]$$, e.g. by proving it goes out for $$\pi/\sqrt{K_0}$$ and goes back by $$\pi/\sqrt{K_0}$$ but I don't see why would that be correct.

What am I missing?

Turns out it was simpler than I thought. Since there exists $$b\in Im(\alpha_{t})$$ such that: $$d(p,b)\geq \frac{\pi}{\sqrt{K}}-\frac{\epsilon}{2}$$ And since both $$\alpha_{t}$$ and $$\gamma_{0}$$ are paths from $$p$$ to $$q$$, we have, from the triangle inequality: $$\ell[\alpha_{t}]+\ell[\gamma_0] \geq d(p,b)+d(b,q)+d(p,q)\geq d(p,b)+d(p,b)\geq \frac{2\pi}{\sqrt{K}}-\epsilon$$