To paraphrase the exercise:
An atlas of class $C^r$ is sometimes defined as a collection of bijective maps from subsets of $X$ to open subsets of $\mathbb R^n$ such that all coordinate changes are $C^r$. Given such an atlas $\Phi=(\phi_i,U_i)_I$, show that there exists a unique topology on $X$ making $\Phi$ a $C^r$ atlas on $X$.
I think that I properly solved this exercise by constructing the minimal topology on $X$ that renders all $\phi_i$ continuous on their given open domains and further showing that this is also the maximal topology for which the charts are bi-continuous.
My question: Is the family $$\left(\phi_i^{-1}(O_i)_{O_i \text{ is open in } \phi_i(U_i)} \right)_{i \in I}$$ a basis for the topology on $X$?
It is clear that every point in $X$ is included in some member of the family but I seem to fail to (explicitly/constructively) prove that for any intersection of two members of a family, there is another member that is included in the intersection.
Remark: Hirsch does not assume any separability or countability axiom for a manifold $M$.
I assume you mean the collection $$\{ \phi_i^{-1}(O) \mid i \in I, O \subset \Bbb R^n \text { is open}\}$$
Suppose $x \in \phi_i^{-1}(O_i) \cap \phi_j^{-1}(O_j)$. Then there is an open neighborhood $N$ of $\phi_i(x)$ such that $\phi_j\circ\phi_i^{-1}$ is a homeomorphism of $N$ to a neighborhood of $\phi_j(x)$.
Let $\tilde N = N \cap O_1 \cap \phi_i \circ \phi_j^{-1}(O_2)$. As the intersection of three open sets $\tilde N$ is open. And $x \in \phi_i^{-1}(\tilde N) \subset \phi_i^{-1}(O_i) \cap \phi_j^{-1}(O_j)$