general-topologygeometric-topologylow-dimensional-topologymanifolds
I would like to know whether the following is true:
Let $M$ be a topological 2-manifold (without boundary), and let $i: [0,1] \to M$ be a continuous embedding. Then $\mathrm{Im}(i)$ is locally flat in $M$.
This question has been asked before, but I could not obtain the statement from the reference given in the comments.
As pointed out there, the counterexamples to local flatness that are commonly given take place one dimension higher.
Let $p \in M$ be a point, and $(U, \varphi)$ be a chart neighborhood in $M$, $\varphi : U \to \Bbb R^n$ being the chart homeomorphism. Considering the embedding $M \subset \Bbb R^k$, find a neighborhood $V$ of $p$ in $\Bbb R^k$ such that $V \cap M = U$ and define a continuous extension $\widetilde{\varphi} : V \to \Bbb R^n$, so that $\widetilde{\varphi}|_U = \varphi$.
Now consider the embedding $M \subset \Bbb R^k \times \{0\} \subset \Bbb R^{k+n}$ into the first factor of $\Bbb R^{k+n} = \Bbb R^k \times \Bbb R^n$ and let $B$ be a ball around $p$ in $\Bbb R^{k+n}$ such that $B \cap (\Bbb R^k \times \{0\}) = V$. Define $\Phi : B \to \Bbb R^{k+n}$ by
$$\Phi(x, y) = (\varphi^{-1}(y + \widetilde{\varphi}(x)) - x, y + \widetilde{\varphi}(x))$$
Checking that $\Phi$ is a homeomorphism onto image is a triviality. Note that for any $(x, 0) \in U \times \{0\} \subset M$, $\Phi(x, 0) = (\varphi^{-1}(\varphi(x)) - x, \varphi(x)) = (0, \varphi(x)) \in \{0\} \times \Bbb R^n$, so $\Phi(U \times \{0\}) \subset \{0\} \times \Bbb R^n$. Thus, $M \subset \Bbb R^{k+n}$ is a locally flat submanifold at $p$.
I've decided to make an edit to explain the idea behind the map $\Phi$. For illustration it's easier to imagine a smaller dimensional example of a non locally flat embedding than the Alexander horned sphere. I shall use a properly embedded wild arc $A \subset \Bbb R^3$. We embed $A$ horizontally in $\Bbb R^3 \times \Bbb R^1$.
Our plan is to shift $A \subset \Bbb R^3 \times \{0\}$ vertically up along the $\Bbb R^1$-direction in $\Bbb R^3 \times \Bbb R^1$ so that it "snakes upward" diagonally as we go along the arc. The natural way to do it is to observe the arc admits a homeomorphism $\varphi : A \to \Bbb R$, which we extend by Tietze extension theorem to a map $\widetilde{\varphi} : \Bbb R^3 \to \Bbb R$, and use this to construct a shear homeomorphism
$$\Phi_1 : \Bbb R^3 \times \Bbb R^1 \to \Bbb R^3 \times \Bbb R^1, \; \Phi_1(x, y) = (x, y + \widetilde{\varphi}(x))$$
The advantage of using $\varphi$ along the arc while shearing is that it "unwinds" the many twists in $A$ because $\varphi$ by definition topologically simplifies the arc to the straight real line.
$\hskip1.4in$ 
What was the point of this construction? The point is now $\Phi_1(A) = \{(x, \varphi(x)) : x \in A\}$ is the graph of $\varphi^{-1} : \Bbb R \to A \subset \Bbb R^3$ over the $\{0\} \times \Bbb R^1$ factor, and graphs of functions are always locally flat submanifolds. Indeed, consider a function $f : \Bbb R^n \to \Bbb R^m$ and $\Gamma_f = \{(x, f(x)) : x \in \Bbb R^n\}$ be the graph in $\Bbb R^n \times \Bbb R^m$. Then another shear homeomorphism
$$\Phi_2 : \Bbb R^n \times \Bbb R^m \to \Bbb R^n \times \Bbb R^m, \; \Phi_2(x, y) = (x, y - f(x))$$
satisfies $\Phi_2(\Gamma_f) \subset \Bbb R^n \times \{0\}$, making $\Gamma_f \subset \Bbb R^n \times \Bbb R^m$ locally flat.
The final homeomorphism $\Phi : \Bbb R^4 \to \Bbb R^4$ such that $\Phi(A) \subset \{0\} \times \Bbb R$, making $A$ locally flat, can be obtained from composing the homeomorphisms $\Phi_1$ and $\Phi_2$, where $\Phi_2$ is done with the Euclidean factors reversed. None of this is special to the example I took, and one can check the formula for $\Phi$ obtained in this manner is exactly what I have above.
One (standard) approach to defining manifolds is to start with a topological space $M$. We say that this topological space is a (topological) manifold if it is (second countable Hausdorff and) is locally homeomorphic to $\mathbb{R}^n$, i.e. every point $p$ in $M$ has an open neighborhood homeomorphic to an open set in $\mathbb{R}^n$. The invariance of domain theorem then implies that 1) for a given $p$ only one $n$ can work and 2) for all $p$ in a given connected component of $M$ this $n$ is the same. So, if $M$ is connected, $n$ is uniquely determined. If not, $n$ is uniquely determined on each component. Most authors then put the requirement that the $n$s for all the components are also equal into the definition of a manifold as well. Such a topological space is then called an $n$-manifold.
Note also that for a topological space being a topological manifold is not a structure, it's a property. It is meaningless to talk about given topological space having two "topological manifold structures". The topological space either is a topological manifold or isn't.
All the extra structures (smooth, complex et cetera) are then on top of this, so of course, they are also all of the same dimension $n$ -- the dimension of the underlying topological manifold $M$.
On the other hand, if you start with a set without any topology then of course any set of the cardinality of the continuum can be topologized to become homeomorphic to any given manifold (and thus can be made the underlying set of a manifold of any dimension bigger than $0$, but for a very silly reason!).
Best Answer
The statement you want does follow immediately from the theorem quoted in the second comment of the link you provided:
Simply use that $i[0,1]$ is homeomorphic to $[0,1]$ and hence abstractly triangulable.