I always had this doubt about the topologies one can introduce into a smooth manifold:
Let's say I have a smooth manifold $M$. We can equip this manifold with a topology induced by the smooth structure as follows: we say that a subset $O\subset M$ is open if and only if $\varphi(O\cap U)$ is an open subset of the euclidean space for each chart $(U,\varphi)$ on the smooth structure. Let's call this topology $\tau_M$.
Conversely, it is usual to start our standard examples of smooth manifolds as topological manifolds (that is, topological spaces which are locally euclidean and Hausdorff, second-countable depending on what kind of property one desires) and equip this manifold with a smooth structure where each admissible chart domain is an open subset of the above topology and the chart maps are homeomorphisms. Let's call this topology $\tau$.
Suppose $M$ is just a topological manifold equipped with a smooth atlas such that each chart domain is an open subset and the chart maps are homeomorphisms as above. In this situation, we can show that the topologies $\tau_M$ and $\tau$ coincide.
For instance, if $O\in\tau$, then $\varphi(O\cap U)$ is open for each admissible chart $(U,\varphi)$ since it is the intersection of both open subsets $O$ and $U$ with respect to $\tau$, hence $O\in\tau_M$. Conversely, if $O\in\tau_M$, then for a given admissible chart $(U,\varphi)$ we have that $\varphi(O\cap U)$ is open, and in particular, $(O,\varphi\lvert_{O\cap U})$ is an admissible chart, and therefore $O=\varphi^{-1}(\varphi(O))$ is open (with respect to $\tau$), since $\varphi$ is an homeomorphism.
However, it is known that we not always can transform a topological manifold into a smooth manifold. So, what am I missing here? What's the obstruction to start with a topological manifold structure and define smooth charts with open domains on this given topology? What is the flaw in the argument above?
Your reasoning seems to be of the form:
Suppose that $(M,\tau)$ is a topological manifold and $A$ is a smooth atlas on $M$ which is ``compatible'' with the topology $\tau$ on $M$. Then $\tau$ is equal to the topology on $M$ induced by $A$.
But if someone hands you a topological manifold $(M,\tau)$ and nothing else, this statement tells you nothing about whether a smooth atlas that is ``compatible'' with $\tau$ exists.