The discussion in the comments is getting too long, let me sum up the proof. This is community wiki; feel free to correct errors and fill in details.
Let $H^*$ denote the Alexander-Spanier cohomology with $\mathbb Z_2$ coefficients. We need the following properties of a compact $n$-dimensional Alexandrov space $X$.
$H^n(X)=\mathbb Z_2$. This is Lemma 1 in Grove-Petersen paper, they also rephrase it as "$X$ has a fundamental class"
For every manifold point $x\in X$, the map $i^*:H^n(X,X-x)\to H^n(X)$ is an isomorphism. This property is referred to as "having a fundamental class that makes sense".
Hopefully this follows from the fact that $X$ has an open dense connected subset which is a manifold and whose complement has dimension at most $n-3$. Someone who knows something about AS cohomology is needed to clarify this.
We also have $H^{n+1}(X)=0$. This and the second property above and the exact cohomology sequence for the pair $(X,X-x)$ imply that
- $H^n(X-x)=0$, if $x$ is a manifold point. (Conversely, the second property above follows from this).
We also need the following special case of Perelman's stability theorem:
- Every point $x\in X$ has a neighborhood homeomorphic to the cone over the space of directions $\Sigma_x$ which is an Alexandrov space of curvature $\ge 1$ and dimension $n-1$, so that the spherical suspension over $\Sigma_x$ is also an $n$-dimensional Alexandrov space. Here $X$ does not need to be compact.
Given all this, the proof works as follows. Let $X,Y$ be $n$-dimensional Alexandrov spaces, $U\subset X$ an open set, $f:U\to Y$ an injective map, $x\in U$ and $y=f(x)$. We are to prove that $f(U)$ contains a neighborhood of $y$. We can make $U$ so small that its closure is compact and $f$ extends to this closure.
Let $U'\subset\subset U$ be a small cone neighborhood of $x$ and $C=U-U'$. Then $U/C$ is homeomorphic to the spherical suspension over $\Sigma_x$ so that it satisfies the above nice properties. Let $D\subset Y$ be a complement of a cone neighborhood of $y$, where the neighborhood is so small that $f(C)$ does not touch it. Now we have a new map $f:U/C\to Y/D$ which is injective on $f^{-1}(Y-D)$. It suffices to show that the new $f$ is onto, and we only need to prove that its image covers all manifold points.
There exists a manifold point $x'\in U-C$ such that $y':=f(x')\in Y-D$ and $y'$ is also a manifold point. By degree theory (which btw depends on domain invariance in $\mathbb R^n$), the map $f^*:H^n(U/C,U/C-x')\to H^n(Y/D,Y/D-y')$ is an isomorphism. By the second property above it follows that $f^*: H^n(U/C)\to H^n(Y/D)$ is an isomorphism.
Now suppose that there exists $z\in Y-D$ which is not in the image of $f$. Then $f$ can be filtered through $Y/D-z$, but $H^n(Y/D-z)=0$, so $f^*: H^n(U/C)\to H^n(Y/D)$ is zero, a contradiction.
