Why 2-tori with Gauss curvature $\geq -1$ cannot collapse to segment? Let $\{(\mathbb{T}^2,g_i)\}_{i=1}^\infty$ be a sequence of 2-dimensional tori with smooth Riemannian metrics with Gauss curvature at least $-1$. It was explained in the final answer to the post Gromov-Hausdorff limits of 2-dimensional Riemannian surfaces that such a sequence cannot converge to a segment in the Gromov-Hausdorff sense.
The argument there used a reduction to the case of collapse of 3-manifolds studied in the paper Shioya–Yamaguchi - Collpapsing three-manifolds under a lower curvature bound.

I would be interested to have a more direct proof of the above fact that 2-tori with Gauss curvature at least $-1$ cannot converge to a segment. 

This fact seems to me to be more elementary than the case of 3-manifolds studied in the above paper. A reference would be very helpful.
 A: This is not quite an answer, bur certainly too long to be a comment! Here are two sketchy ideas.
By following the arguments of Shioya and Yamaguchi, you could make rigorous this argument that the only orientable 2-manifold collapsing to an interval is $S^2$. The result will not be so elementary as an application of the Gauss--Bonnet Theorem, but it is surely more straightforward than going up to dimension three.
The portion of $M^2$ which is collapsing to the interior of the interval fibers over it, so is homeomorphic to $S^1 \times I$. Now you need to cap this off with two pieces corresponding to the ends of the intervals. Let $p_i$ be a sequence of points converging to an end of the interval. Rescale the sequence $(M^2, p_i)$ so that you obtain a non-negatively curved 2-dimensional limit space, which can have only one end. This is either $D^2$ or the Möbius strip -- by orientability it must be $D^2$. Therefore $M^2$ is given by $S^2$.
Something a little more elementary might be as follows. Suppose $M^2$ were a torus. Cut the interval in two and lift this cut through the fibration to divide $M^2$ along a circle. This describes $M^2$ as a connected sum, and the only way to do that with a torus is as $T^2 \# S^2$. Carefully piece two copies of the $(T^2 \setminus D^2)$ portion together preserving the lower curvature bound -- fiddly, but almost certainly doable. In this way we obtain a genus two surface collapsing to an interval and so a contradiction.
