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 $D$ 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 $D/10$.

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 $D/10$.