Let us first review our topological notions to ensure there isn't any confusion there.
A topological space is a set $X$ together with a collection $U$ of subsets of $X$, called a topology on $X$, so that the following hold,
- The empty set $\varnothing$ and the set $X$ are both open.
- $U$ is closed under arbitrary union: $$\text{If } U_{1},...,U_{k} \text{ are open, then } \bigcup_{i =1}^k U_{i} \text{ is open.}$$
- $U$ is closed under finite intersections: $$\text{If } U_{1},...,U_{k} \text{ are open, then } \bigcap_{i=1}^k U_{i} \text{ is open.}$$
We can define maps between topological spaces $X$ and $Y$ where,
A map $f: X \rightarrow Y$ is continuous if $f^{-1}(U)$ is open for every open set $U \subset Y$. A continuous map that has a continuous inverse is a homeomorphism and if there exists a homeomorphism between $X$ and $Y$, then $X$ and $Y$ are considered topologically equivalent.
Generally, a topological manifold can be thought of as some space which locally looks like $\mathbb{R}^n$ endowed with a metric by which calculus can be performed. More precisely, a topological space $\mathcal{M}$ is locally Euclidean of dimension $n$ if and only if every point of $\mathcal{M}$ has a neighbourhood that is homeomorphic to an open subset of $\mathbb{R}^n$. We then have our initial definition of a topological manifold as follows,
A topological space $\mathcal{M}$ is called a topological $n$-manifold when,
$\mathcal{M}$ is locally Euclidean of dimension $n$; that is, for each $p \in \mathcal{M}$ there exists,
- An open set $U \subset \mathcal{M}$ with $p \in U$,
- An open set $\widetilde{U} \subset \mathbb{R}^d$ such that $\varphi : U \rightarrow \widetilde{U}$ is a homeomorphism.
Furthermore, $\mathcal{M}$ is a Hausdorff space: For every two distinct points $p,q \in \mathcal{M}$, there are disjoint open subsets $U,V \subset \mathcal{M}$ such that $p \in U$ and $q \in V$. An additional property that $\mathcal{M}$ is second countable (there is a countable basis for the topology of $\mathcal{M}$) gives a more general definition
Examples of topological manifolds, are $\mathbb{R}^n$, the unit sphere $\mathbb{S}^n$, graphs of continuous functions $f: U \rightarrow \mathbb{R}^d=n$, and the real projective plane $\mathbb{RP}^n$.
In general the reason for defining a topological $n$-manifold with the additional characteristics to locally Euclidean is to avoid certain pathological examples that we do not want to count as a manifold. For example, to explain why we would want a space to be Hausdorff, I'll present the best illustration I have found so far: we want all points to be, in a sense, separable. Consider the real line $\mathbb{R}$ glued to another real line $\mathbb{R}'$ along the open sets $(-\infty,0) \cup (0,\infty)$ away from the origin. Then we cannot embed this space in Euclidean space because there are multiple ways to fill in a jump discontinuity of a function, and compact sets are not necessarily closed.
For a space to be second-countable we need to avoid the uncountable well-ordered sets guaranteed to exist by Zorn's Lemma. Firstly recall that if $(X,<)$ is a partial order, then $C \subseteq X$ is a chain in $X$ if $C$ is linearly order by $<$.
That is, Zorn's Lemma states,
If $(X,<)$ is a partial order such that for every chain $C \subseteq X$ there is $x \in X$ such that $c \leq x$ for all $c \in C$, then there is $y \in X$ such that the is no $z \in X$ with $z > x$. In other words, if every chain has an upper bound, then there is a maximal element of $X$.
From this we can give a proof of the well-ordering principle which states any there is always a well-ordering of $A$ (for any nonempty $C \subseteq A$, there is an $a \in C$ such that $a \leq b$ for all $b \in C$). Now, the actual reason we need to have our manifold be second countable, is that Zorn's lemma guarantees the existence on an uncountable well-ordered set. So, given the order topology we can take half-open intervals indexed by our uncountable well-ordered set and have that the space cannot be embedded in $\mathbb{R}^n$ and that there is no countable dense subset (not separable). So, we require our space to be second countable because we want to avoid such pathological examples to do with bizzare uncountable orderings that lose analytic separability.
I will stop rambling now, but hopefully this is helpful; the general idea that I would take away is there there are good reasons for defining a topological $n$-manifold the way it is commonly defined because there are certain "counterexamples" or pathological examples that do not have the nice properties. If we are to classify such a vast collection of spaces, as "all topological manifolds of any dimension", we certainly what the power of generality to not have to deal with pathological examples that cause general theorems to not hold because of some bizzare uncountable ordering or the like.