Concluding that $\mathbb{R}P^2$ cannot be embedded in $S^3$ I want to deduce that the projective plane $\mathbb{R}P^2$ cannot be embedded in $S^3$ using homology. My idea is to compute $H_*(S^3-h(\mathbb{R}P^2))$ for some arbitrary embedding $h:\mathbb{R}P^2\to S^3$ and obtain a contradiction. 
Previously I've computed $H_*(S^3-h(M))$, where $M$ is the Möbius strip, which is $\mathbb{Z}$ for $*=0,1$ and $0$ otherwise, and I also have that the homomorphism $i_*:H_1(S^3-h(M))\to H_1(S^3-h(\partial M))$ induced by inclusion is multiplication by $2$
My work
Now, I know that $\mathbb{R}P^2=M\cup int D^2$, so I can write $S^3-M=(S^3-\mathbb{R}P^2)\cup intD^2$. I have $(S^3-\mathbb{R}P^2)\cap intD^2=\emptyset$. This set theoretic relation is preserved by $h$, so the Mayer-Vietoris sequence tells me immediately that $H_i(S^3-h(\mathbb{R}P^2))=0$ for $i\geq 2$. For $i=1$ it is also easy since we have
$0\to H_1(S^3-h(\mathbb{R}P^2))\oplus H_1(h(intD^2))\to H_1(S^3-h(M))\to 0$
Since $h(int D^2)$ is contractible, we get $H_1(S^3-h(\mathbb{R}P^2))\cong H_1(S^3-h(M))=\mathbb{Z}$. Similarly, for $i=0$ 
$0\to H_0(S^3-h(\mathbb{R}P^2))\oplus H_0(h(intD^2))\to H_0(S^3-h(M))\to 0$
Since both $h(intD^2)$ and $S^3-h(M)$ are path connected, this forces $H_0(S^3-h(\mathbb{R}P^2))=0$. The contradiction now is that this means that $S^3-\mathbb{R}P^2$ has no path-connected components so it must be empty. This implies $S^3=h(\mathbb{R}P^2)$, which is not possible since $S^3$ is not homeomorphic to $\mathbb{R}P^2$.
Question
Is everything okay? I'm unsure whether there is a step where I missed something, especially the contradiction part. I didn't use $i_*$, so if there is an alternative proof using it I'll appreciate it.
 A: For a cleaner and more general argument we can use Alexander Duality: 

Let $X$ be a compact, locally contractible subspace of the sphere $S^{n}$ of dimension $n$. Let $S^{n}\setminus X$ be the complement of $X$ in $S^{n}$. Then if $\tilde {H}$ stands for reduced homology or reduced cohomology, with coefficients in a given abelian group, there is an isomorphism
  $$\tilde {H}_{q}(X)\cong \tilde {H}^{n-q-1}(S^{n}\setminus X)$$
  for all $q\geq 0$. Note that we can drop local contractibility as part of the hypothesis, if we use Čech cohomology, which is designed to deal with local pathologies.

(EDIT: My original answer was for the case $n=3$, but it immediately generalized to higher dimensions as well.)
A smooth embedding $X\to S^n$ where $X$ is a closed manifold will have the required properties. In this case Alexander Duality provides a generalization of a Jordan Separation theorem:
Corollary: If $X$ is a closed, connected manifold of dimension $n-1$ and $X \to S^{n}$ is a smooth embedding then $S^{n} \setminus X$ has two components.
Proof: Recall that for any $Y$, $rank_{\mathbb{Z}/2}H^0(Y;\mathbb{Z}/2)$ counts the path components of $Y$, and reduced cohomology gives you one less than that. Recall that $H_{n-1}(X;\mathbb{Z}/2)\cong \mathbb{Z}/2$ regardless of wether it is orientable or not. Then by Alexander Duality$$ \mathbb{Z}/2 \cong H_{n-1}(X;\mathbb{Z}/2) \cong \tilde{H}^0(S^{n} - X;\mathbb{Z}/2).$$
Corollary:
For $X$ as above, if there is an embedding $X\to S^n$ then $X$ must be orientable.
Proof: Now we take integer coefficients, and use the fact that $X$ is orientable iff $H_{n-1}(X;\mathbb{Z})\cong \mathbb{Z}$. Again by Alexander Duality
$$ H_{n-1}(X) \cong \tilde{H}^0(S^n - X)\cong \mathbb{Z}$$
where the last isomorphism holds because of the claim.

Now we can conclude that $\mathbb{RP}^{2n}$ does not embed into $S^{2n+1}$ because it is not orientable. 
On the other hand $\mathbb{RP}^{2n+1}$ is orientable so another argument would be needed if you wanted to show non-embedability (and in fact $\mathbb{RP}^1\cong S^1$ does embed into $S^2$).
