Suppose $X\subseteq\mathbb{R}^m$ s.t. for any $x\in X$ and any open $U\subseteq\mathbb{R}^m$ that contains $x$, there exists a smaller open set $V\subseteq U$ also containing $x$, so that $V\cap X$ is the image of some injective continuous map $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$.
Any $n$-manifold in $\mathbb{R}^m$ clearly satisfies this property; conversely must such an $X$ be an $n$-manifold? At least I cannot come up with a $1$-dimensional counterexample; the usual examples such as the topologist's sine curve are ruled out by this definition, since it implies locally connectedness. It also rules out the figure 8 since the X shape cannot be injective image of $\mathbb{R}$.
I am asking in the continuous category, but feel free to assume smoothness for simplification.
Clarification: at first I just let $X$ be any topological space, so people thought I was looking for examples of non-standard manifolds; also someone pointed out that $\mathbb{R}^n$ with discrete topology is a counterexample. Therefore I decided to require $X$ to be a subset of some Euclidean space, but this is really the case which I think is more interesting.
Update: I feel like at least this is true for $n=1$. Here is an outline, although I'm not confident about any of these steps.
(i) Show that if $Y\subseteq X$ is the injective image of $\mathbb{R}$, then it cannot be the injective image of disjoint copies of $\mathbb{R}$.
(ii) If $f,g$ are both injective map from $\mathbb{R}$ to $X$ and $f(\mathbb{R})\cap g(\mathbb{R})\neq\emptyset$, then $f(\mathbb{R})\cup g(\mathbb{R})$ is either also such an image or a circle.
(iii) Each point of $X$ is contained in either a circle or the image of some injective $f:\mathbb{R}\rightarrow X$ that is maximal (no larger set is such an image).
(iv) Conclude that $X$ is a $1$-manifold.