Generalizing a theorem from real analysis regarding strictly increasing functions

This is just horribly false. For instance, even in just the case where the domain is an interval $[a,b]$, consider the map $f:[0,2\pi]\to S^1$ given by $f(t)=e^{it}$. Then $f$ is an embedding on $(0,2\pi)$, but $f$ is not injective.

It's not even true if you restrict to (say) open subsets of $\mathbb{R}^n$ and maps to $\mathbb{R}^n$. For instance, the map $f:[0,1]^2\to\mathbb{R}^2$ given by $f(x,y)=(x,xy)$ is an embedding on its interior (an open subset of $\mathbb{R}^2$) but is not injective on the boundary since it collapses the entire boundary edge $\{0\}\times[0,1]$ to a point.

Your argument goes wrong when it asserts that the image of $S_p$ is a neighborhood of $f(p)$. That's not necessarily true (indeed, in the example above, the image of $[0,1)$ is not a neighborhood of $f(0)$, for instance).

I don't know if I am missing something but your statement looks wrong. $e^{ix}$ is a continuous map from $[0,2\pi]$ into $S^{1}$ and it is a homeomorphism om $(0,2\pi)$. But this map is not injective on $[0,2\pi]$.