Riemannian metrics on 2-sphere invariant under antipodal involution EDIT: Let $g$ be a smooth Riemannian metric on $S^2$ which is invariant under the antipodal involution $x\mapsto -x$. Assume it has Gauss curvature at least $-1$ and diameter $D$. 

Does there exist a constant $c(D)>0$ (but independent of the metric) such that $\sup_x dist(x,-x)\geq c(D)$? 

 A: The answer is "yes".
Let $g$ be the induced Riemannian metric on $\mathbb R\mathrm P^2=\mathbb S^2/\mathbb Z_2$.
Denote by $\ell$ the systole of $(\mathbb R\mathrm P^2,g)$ and let $\gamma$ be the corresponding closed geodesic.
Let us cut $(\mathbb R\mathrm P^2,g)$ along $\gamma$. 
We obtain a disc $\Delta$ with diameter is at least $D/2$ and its boundary has length $2\cdot\ell$.
Note that we can assume that $\ell<D/100$.
Therefore there is a point $x$ in $\Delta$ that lies on distance at least $D/10$ from the boundary.
It remains to note that the distance from $x$ to its antipodal point is more than $D/10$.
A: This is not an answer. Following the request of Igor Belegradek I would like to elaborate on Anton Petrunin's answer and prove that $diam (\Delta) \geq D/2$.
First of all $diam (\mathbb{R}\mathbb{P}^2)\geq D/2$. This is a special case of the more general estimate proved in the final answer to this post:   Diameter of m-fold cover
Let $f\colon S^2\to \mathbb{R}\mathbb{P}^2$ be the canonical map. Let $a\colon S^2\to S^2$ be the antipodal involution. Then $\tilde\gamma:=f^{-1}(\gamma)$ is a closed connected (!) geodesic on $S^2$, it is $a$-invariant. By the Jordan theorem its complement consists of two disks whose closures will be denoted by $\Delta $ and $\Delta'$. (Say, the first disk is the one Anton constructed, I guess.) Then
$$a(\Delta)=\Delta'.(1)$$
Indeed otherwise $a(\Delta)=\Delta$. In that case the group $\mathbb{Z}_2$ would act freely on the disk $\Delta$ which is imposible since in that case the quotient $\Delta/\mathbb{Z}_2$ would be a compact manifold (possibly with boundary) of Euler characteristic 1/2 which is absurd.
Let $x,y\in \mathbb{R}\mathbb{P}^2$ be such that $dist(x,y)\geq D/2$. Let $\tilde x,\tilde y \in \Delta$ be their lifts (they can be chosen to belong to $\Delta$ due to (1)).
Then one has
$$diam(\Delta)\geq dist(\tilde x,\tilde y)\geq \min\{dist(\tilde x,\tilde y),dist (\tilde x,a(\tilde y))\}=dist(x,y)\geq D/2.$$
QED
