Let $G=\text{SL}(d,\mathbb R)$ and $\Gamma=\text{SL}(d,\mathbb Z)$ and $X=G/\Gamma$. (Maybe we only need $\Gamma$ to be discrete I don't know.)
Consider a convergent sequence $g_i\Gamma \to g\Gamma$. I wonder if we can rechoose the representatives of cosets (for both $g_i\Gamma$ and $g\Gamma$) so that $$g_i\to g$$
For this special $G$ and $\Gamma$ I think I have a proof and here is a sketch: I need the possibly unnecessary condition that $\Gamma$ is discrete and $G/\Gamma$ is metrizable. First observe that $g_i\Gamma \to g\Gamma$ is equivalent to $g^{-1}g_i\Gamma$. We choose $\gamma_i$ to be the element in the (closed) subgroup $\Gamma$ of shortest distance to $g^{-1}g_i$, then $g^{-1}g_i \gamma_i^{-1} \to e$, or $g_i\gamma_i^{-1} \to g$. But this might be true for more general $G$ and $\Gamma$.