why can't a topological manifold have a finite underlying set? According to a course notes I am following, a topological space cannot have a finite underlying set, because it is locally homeomorphic to $R^d$, and $R^d$ is an uncountable set.
However, why can't I define the set $M=$ { $1$ } together with the indiscrete topology $O$, and the map $f: M \to R$, where $f(1)= e$ (or any other real number), to be a finite topological space that maps to a single element in $R$? 
In this case there are no open subsets of $R$ to which $f$ maps, so the preimage of all open maps in the target is the nullset in the domain, which is also an open set in the domain. 
Hence $(M, O, f)$ should be a topological manifold. Anything wrong with my reasoning?
 A: The definition of a "chart" or "coordinate patch" on the manifold $M$ is a homeomorphism $h \colon U \to V$ from some open subset $U \subset M$ to an open subset $V \subset \mathbb R^d$. That means $V$ must be open as a subset of  $\mathbb R^d$ rather than just being open as a subset of itself under the subspace topology. The range of your example is not an open subset of $\mathbb R$.
A: The map $f$ you defined in your Question is not an homeomorphism to $R= R^1$, that's why your idea doesn't work. (For instance it is not surjective; but you could work with a more general definition that every point has a neighbourhood that is homeomorphic to an open subset of some $R^n$. But in that case the image of your $f$ is not open in $R$)
However a disjoint union of finitely points (all closed as subset in the space) are a manifold, as every point $x$ has an open neighbourhood (namely $\{x\}$) that is homeomorphic to $R^0$ wich is the $0$-dimensional euclidean space. In general you don't want to exclude this case, and I believe your definition doesn't do that aswell.
