# What does May mean by "inductively" choosing paths in van Kampen's proof?

I am reading the fundamental group (not groupoid!) version of May's categorical proof van Kampen'd theorem. I am having trouble understanding what he means by "inductively" choosing paths here:

The inclusion of categories $$J \colon \pi_1(X,x) \to \Pi(X)$$ is an equivalence. An inverse equivalence $$F : \Pi(X) \to \pi_1(X,x)$$ is determined by a choice of path classes $$x \to y$$ for $$y \in X$$; we choose $$c_x$$ when $$y=x$$ and so ensure that $$F\circ J=\mathrm{Id}$$. Because the cover $$\mathscr O$$ is finite and closed under finite intersections, we can choose our paths inductively so that the path $$x \to y$$ lies entirely in $$U$$ whenever $$y$$ is in $$U$$.

I am not able to understand what is being done in the statement that is bolded out. All I am getting is that the paths $$\gamma_y\colon x\to y$$ are being chosen so that $$\gamma_y$$ lies entirely in some $$U$$, but this doesn't use finiteness of $$\mathscr O$$ for each $$U\in\mathscr O$$ is path-connected. I surely am missing something essential here, something relating to "inductively" and the fact that $$\mathscr O$$ is finite. Can someone nudge me towards the correct thing?

May chooses the path $$\gamma_y\colon x\to y$$ such that for any $$U\in \mathcal{O}$$ that contains $$y$$, $$\gamma_y$$ lies in $$U$$. To do this, just take the intersection of all $$U\in \mathcal{O}$$ that contain $$y$$. By the assumptions this intersection is again in $$\mathcal{O}$$ and since $$\mathcal{O}$$ consists of path connected sets, it contains a path from $$x$$ to $$y$$.