Here an outline of an argument for an affirmative answer in the case n=1:
By replacing $X$ by $X\cap V$, we can assume $X=g(\mathbb R)$ for a continuous injective map $g:\mathbb R\rightarrow \mathbb R^m$.

We prove (under the given assumptions on X) that $g$ is a proper map to its image (and thus a homeomorphism to its image). Since $g$ is locally on $\mathbb R$ a homeomorphism to its image, it suffices to show that no $x\in X$ is the limit of a sequence $g(t_k)$
with $|t_k|\rightarrow \infty$.

Assume by contradiction, that there is $x\in X$ with $g(t_k)\rightarrow x$ for some $t_k\rightarrow \infty$.

(*) Assume further for simplicity that $g(t)$ converges to $z\neq x$ at the other end of $\mathbb R$, as $t\rightarrow -\infty$.

- We will show that in this situation $\lim_{t\rightarrow \infty}g(t)=x$.

Indeed, if there is another accumulation point $y\in \mathbb{R}^m$ of $g(t)$ as $t\rightarrow \infty$, pick an open neighborhood $V$ of $x$ such that $y,z\notin \overline{V}$.

Then $g^{-1}(V)$ consists of a countably infinite disjoint union of nonempty open intervals $I_k\subset \mathbb R$ with compact closures (these intervals tend to $\infty$, since $g$ only accumulates to $x$ as $t\rightarrow \infty$). The images of $J_k:=g(I_k)\subset V$ are homeomorphic to open intervals. The closures of the $J_k$'s in $\mathbb R^m$ are pairwise disjoint and their boundary $\overline{J_k}\setminus J_k\subset \partial V$, consists of two points for each $k$. Thus each $J_k$ is *closed* in $V$.

Assume now $V$ is also such that $X\cap V$ is the image of a continuous injective map $f:\mathbb R\rightarrow \mathbb R^m$. Since we can write $X\cap V= \bigsqcup_k J_k$ as a disjoint union of countably many nonempty closed subsets, the properties of $f$ imply that $\mathbb R$ is the disjoint union of countably many nonempty closed subsets. But this is not possible (e.g. by an application of the Baire category theorem), thus we find a contradiction.

- Obtain from 1. and (*) a contradiction to our assumptions on $X$:

$g(t)$ converges to $x$ as $t\rightarrow \infty$, but does not accumulate at $x$ for $t\rightarrow -\infty$. Thus $X\setminus {x}$ has locally near $x$ 3 connected components, which shows that $X\setminus{x}$ locally near x cannot be the continuous image of $\mathbb{R}\setminus p$. This contradicts the assumptions on $X$.

- Removing the assumption $(*)$.
There are several cases to consider:

a) If $x$ is not an accumulation point of $g(t)$ as $t\rightarrow -\infty$
then the argument from 1. goes through by choosing $V$ so small that it contains no such accumulation point, e.g. such that $g((-\infty,0])\cap V=\emptyset$ (and by ignoring $z$). Step 2. applies now as before to yield a contradiction to the assumptions on $X$.

b) If $x$ is an accumulation point of $g(t)$ as $t\rightarrow -\infty$ but not the limit of $g(t)$ as $t\rightarrow -\infty$, then the argument from 1. still goes through to derive a contradiction, if we choose $z$ as another accumulation point of $g$ as $t\rightarrow -\infty$ (there are now intervals $I_k$ tending to both $\pm \infty$). This contradiction obtained as in 1. shows (without any further assumptions) that no $x\in X$ can be "accumulation point but not limit" at both ends of $\mathbb R$. Up to switching $\pm\infty$, we are thus in the situation c) if $(*)$ does not hold:

c) $x$ is the limit of $g(t)$ as $t\rightarrow -\infty$ (and an accumulation point, but not the limit, of $g(t)$ as $t\rightarrow \infty$).
In this situation, we still follow the basic strategy of 1., picking V and z as before, but the argument changes as slightly: The connected components of $g^{-1}(V)$ can be enumerated as follows. One component has the form $I_0:=(-\infty,a)$, one (potentially different) component is an open interval $I_0'\subset \mathbb R$ containing $g^{-1}(x)$, and the other components form a countable sequence of pairwise disjoint open intervals $I_k, k\geq 1$ tending to $\infty$ (and not containing $x$). The images $g(I_k), k\geq 1$ are closed in V as before, but also $g(I_0\cup I_0')$ is closed in V (since the only accumulation point of $g(I_0)$ in $V\setminus g(I_0)$ is $x\in g(I_0')$) and all of these sets are pairwise disjoint. Thus one obtains a contradiction as in 1., and it follows that $x$ is the limit of $g$ at both $\pm \infty$.
Then $X\setminus x$ has (similar as in 2.) locally near $x$ 4 connected components, and thus it cannot be the continuous image of $\mathbb R\setminus p$.

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