This can be done with a few transversality arguments. Most of the relevant background information can be found in chapter $6$ of Lee's Introduction to Smooth Manifolds. While this is by no means the most eloquent argument, (my choices of parameterized families are probably overkill), your claim appears to be true for $n\ge 2m+1$.
Throughout, $M$ is a smooth compact manifold without boundary of dimension $m$, $TM$ is its tangent bundle, and $SM\subseteq TM$ be an arbitrary choice of sphere bundle (the submanifold of tangent vectors which have unit norm w.r.t. some Riemannian metric).
The main theorem is the parametric Transversality theorem, which I'll quote from Lee.
Parameteric Transversality Theorem [Lee, Thm 6.35] Suppose $N$ and $M$ are
smooth manifolds, $X\subseteq M$ is an embedded submanifold, and $\{F_s:s\in S\}$ is a
smooth family of maps from $N$ to $M$. If the map $F:N\times S\to M$ is transverse
to $X$, then for almost every $s\in S$, the map $F_s:N\to M$ is transverse to $X$.
Two other useful theorems:
Whitney Approximation Theorem: For any continuous map $f:M\to\mathbb{R}^n$ and any $\epsilon>0$, there is a smooth map $g:M\to\mathbb{R}^n$ which is $\epsilon$-close to $f$.
Whitney Immersion Theorem: There exists a smooth immersion $f:M\to\mathbb{R}^n$ for $n\ge2m$.
In addition to these, some other lemmas are required, which I haven't been able to find anywhere (though the first is an exercise in Lee).
Lemma 1: For any smooth map $f:M\to \mathbb{R}^n$ with $n\ge 2m$, and any $\epsilon>0$, there is an immersion that is $\epsilon$-close to $f$.
Proof: Fix $f,\epsilon$ as above, and let $\varphi:M\to\mathbb{R}^n$ be a Whitney immersion. By composing with a smooth map $\mathbb{R}^n\to\mathbb{B}^n$, we can assume that $\operatorname{im}(\varphi)$ is contained in the unit ball. Let $f_A$ be defined by
$$
f^i_A(x)=f^i(x)+A^i_j\varphi^j(x)
$$
Where $A\in\mathbb{R}^{n\times n}$. Let $F:SM\times\mathbb{R}^{n\times n}\to\mathbb{R}^n$ be a parameterized family of maps defined by $F(v,A)=d_xf_A(v)$, where $v\in S_xM$. I claim that $F$ is transverse to $\{0\}\subset\mathbb{R}^n$. To see this, suppose $F(v,A)=0$. Since $\varphi$ is an immersion and $v\neq 0$, $d_x\varphi(v)\neq 0$. Fixing $i$ and choosing $B\in\mathbb{R}^{n\times n}$ such that $B^j_kd_x\varphi^k(v)=\delta^j_i$ (such a $B$ always exists), we have
$$
\frac{d}{dt}\left(F(v,A+tB)\right)|_{t=0}=B^j_kd_x\varphi^k(v)\partial_j=\partial_i
$$
Since this can be done for any $i\in\{1,\dots,n\}$, and the $\partial_i$ form a basis of $\mathbb{R}^n$, $dF$ is surjective, and thus $F$ is transverse to $\{0\}\subset\mathbb{R}^n$. Now by applying parametric transversality theorem, $df_A:SM\to\mathbb{R}^n$ is transverse to $0$ for almost all $A\in\mathbb{R}^{n\times n}$. Since $SM$ has dimension $2m-1$, this means that for $n\ge 2m$, $0\notin df_A(SM)$, and thus $df_A$ is full rank and $f_A$ is an immersion for almost all $A$.
The set $S=\{B\in\mathbb{R}^{n\times n}:\|B\|_{op}<\epsilon\}$ is open, so there is an $A\in S$ such that $f_A$ is an immersion. Since $\|\varphi(x)\|\le 1$, we have
$$
\|f_A(x)-f(x)\|=\|A\varphi(x)\|<\epsilon
$$
And thus $f_A$ is $\epsilon$-close to $f$.
$\square$
Lemma 2: For any smooth immersion $f:M\to\mathbb{R}^n$ with $n\ge 2m+1$, and any $\epsilon>0$, there is an injective immersion that is $\epsilon$-close to $f$.
Proof: Since $f$ is an immersion, it is a local embedding, and so we may choose a finite covering $\{U_\alpha\}_{\alpha=1,\dots,k}$ of $M$ by extendable open balls $U_\alpha$ such that $f|_{\overline{U_\alpha}}$ is injective for each $\alpha$, and there is a collection of bump functions $\psi_\alpha:M\to\mathbb{R}$ such that $\psi_\alpha|_{U_\alpha}>0$ and $\psi_\alpha|_{M\setminus U_\alpha}=0$.
Let $S=\{(x,y)\in M^2:x\neq y\}$, and define $\Delta_f:S\to\mathbb{R}^n$ by $\Delta_f(x,y)=f(x)-f(y)$. Note that $f$ is injective iff $\Delta_f$ is nonvanishing.
To find a nearby injective map, we can define a parametric family $F:M\times \mathcal{V}\to\mathbb{R}^n$, by
$$
F^i(x,A)=f^i(x)+\sum_{\alpha=1}^kA^i_\alpha\psi_\alpha(x)
$$
Where $\mathcal{V}\subset\mathbb{R}^{n\times k}$ is a neighborhood of zero such that $F_A|_{\overline{U_\alpha}}$ is injective for all $\alpha\in\{1,\dots,k\}$, $A\in\mathcal{V}$. We define $\Delta_F:S\times\mathbb{R}^{n\times k}\to\mathbb{R}^n$ by $\Delta_F((x,y),A)=F(x,A)-F(y,A)$. To show that $\Delta_F$ is transverse to $0$, we choose $(x,y)\in S$ such that $\Delta_F(A,(x,y))=0$. Choosing $\alpha,\beta$ such that $x\in U_\alpha,y\in U_\beta$, it must be the case that $x\notin U_\beta$ and $y\notin U_\alpha$. Thus, choosing $B\in\mathcal{V}$ such that $B^i_\alpha\neq0$ and all other entries are zero,
$$
\frac{d}{dt}(\Delta_F((x,y),A+tB)|_{t=0}=B^j_\alpha\psi_\alpha(x)\partial_j=c\partial_i,\ \ \ c\neq 0
$$
Again doing this for $i\in\{1,\dots,n\}$ gives a basis for $\mathbb{R}^n$, ad so $d\Delta_F$ is surjective wherever $\Delta_F$ vanishes, and so $\Delta_F$ is transverse to $\{0\}\subset\mathbb{R}^n$. By parametric transversality theorem, $\Delta_{F_A}$ is transverse to zero for almost all $A\in\mathcal{V}$, and thus if $n\ge 2m+1$, $F_A$ is injective for almost all $A\in V$. We may thus choose an $A$ such that $F_A$ is injective and $\epsilon$-close to $f$.
$\square$
Combining these two lemmas and the Whitney approximation theorem, we see that for any continuous map $f:M\to\mathbb{R}^n$ with $n\ge 2m+1$, there is an embedding which is $\epsilon$-close to $f$. This bound is sharp. For $n=2m$, immersions can intersect themselves transversally, and these intersections will be stable w.r.t. $L^\infty$-small deformations (consider for instance the $\infty$ symbol as a map $S^1\to\mathbb{R}^2$).
It seems likely that this remains true for other, finer topologies, such as Sobolev or Fréchet topologies on $C^k(M,\mathbb{R}^n)$. The proofs would be a bit more nuanced, of course.
