Let $\{M_i\}$ be a sequence of 2-dimensional orientable closed surfaces of genus $g$ with smooth Riemannian metrics with the Gauss curvature at least $-1$ and diameter at most $D$. By the Gromov compactness theorem, one can choose a subsequence converging in the Gromov-Hausdorff sense to a compact Alexandrov space with curvature at least $-1$ and Hausdorff dimension 0,1,or 2. One can show (see below) that if $g\geq 2$ then the limit space cannot be a point, thus the dimension of the limit space is at least 1 (while for $g=0,1$ it can be 0). 

Let us assume that the limit space has dimension 1. Then it is either circle or segment. 

**Whether these both possibilities (circle and segment) can be obtained in the limit?** 

**ADDED:** It is not hard to see that one can get segment for $g=0$ and circle for $g=1$. I suspect (but cannot prove) that for $g\geq 2$ and $g=0$ one *cannot* get circle in the limit. In fact I do not even know whether in the case $g\geq 2$ a 2-dimensional limit is the only possibility.

**UPDATE:** Based on the answer by Igor Belegradek, let me summarize the situation. Let $\{M_i\}$ be a sequence of genus $g$ orientable closed surfaces with Riemannian metrics with Gauss curvature at least -1 which converges in the Gromov-Hausdorff sense to an Alexandrov space $X$. 

1) If $g=0$ then $X$ is either a point, or a segment, or $X$ is homeomorphic to $S^2$ (by Perelman stability theorem), and all the three cases are possible.

2) If $g=1$ then $X$ is either a point, or a circle $S^1$, or homeomorphic to the 2-torus, and all the three cases are possible.

3) If $g\geq 2$ then $\dim X=2$ and hence $X$ is homeomorphic to an orientable genus $g$ closed surface.

**ADDED:** Let me add a proof that if $g\geq 2$ then a point cannot be the limiting space. Indeed otherwise we would have $d_i:=diam(M_i)\to 0$. Let us divide the metric of $M_i$ by $d_i$ and denote the new metric space by $N_i$. Then the sectional curvature of $N_i$ is at least $-d_i^2$ and diameter 1.

By the Gauss-Bonnet $$4\pi(1-g)=\int_{N_i}K\geq -d_i^2vol(N_i).$$
By the Bishop inequality $vol(N_i)$ is bounded from above. Hence the right hand side in the above inequality tends to 0. Hence $1-g\geq 0$ which is a contradiction.