Any countable Hausdorff space $Q$ is totally path-disconnected. Indeed, if $f:[0,1]\to Q$ is continuous, then its image $X$ is a countable connected compact Hausdorff space. By Urysohn's lemma, then, continuous maps from $X$ to $[0,1]$ separate points. But $X$ is connected, so the image of a continuous map from $X$ to $[0,1]$ is connected, and so must be just a single point since $X$ is countable. Thus $X$ can only have one point, so $f$ is constant.

So, in particular, $\mathbb{Q}\mathbb{P}^\infty$ is totally path-disconnected, and has the weak homotopy type of a countable discrete space.

Any countable Hausdorff space $Q$ is totally path-disconnected. Indeed, if $f:[0,1]\to Q$ is continuous, then its image $X$ is a countable connected compact Hausdorff space. By Urysohn's lemma, then, continuous maps from $X$ to $[0,1]$ separate points. But $X$ is connected, so the image of a continuous map from $X$ to $[0,1]$ is connected, and so must be just a single point since $X$ is countable. Thus $X$ can only have one point, so $f$ is constant.

So, in particular, $\mathbb{Q}\mathbb{P}^\infty$ is totally path-disconnected, and has the weak homotopy type of a countable discrete space.

(In fact, more strongly, any countable $T_1$ space is totally path-disconnected. See Why are the integers with the cofinite topology not path-connected?)