Why does $\mathbb{RP}^2$ not continuously embed in $\mathbb{R}^3$? Ok, I know the answer: any closed hypersurface of $\mathbb R^3$ is orientable while $\mathbb R \mathbb P^2$ is not. 
But I know how to prove that only for smooth embeddings. Is there a simple way to prove that there is no continuous embedding of the projective plane in Euclidean tridimensional space?
 A: The answers I've found in MathSE (1, 2) indeed appear to assume smoothness. 
Let $A \subset \mathbb{R}^3$ be the embedding of a compact hypersurface on $\mathbb{R}^3$. Since it is embedded in $\mathbb{R}^3$, we can also embed it in $S^3$. By Alexander-Poincaré duality, $\check{H}^{2}(A;\mathbb{Z}) \simeq H_1(S^3,S^3-A;\mathbb{Z})$, and this latter one is isomorphic to $\widetilde{H}_0(S^3-A;\mathbb{Z})$. (This can be seen by looking at the reduced exact sequence of the pair $(S^3,S^3-A)$.) Therefore, $\check{H}^{2}(A;\mathbb{Z})$ must be a free $\mathbb{Z}$-module.
Once we know that $\check{H}^{2}(A;\mathbb{Z}) \simeq H^{2}(A;\mathbb{Z})$, the result then follows from the fact that $H^{2}(\mathbb{R}P^2;\mathbb{Z}) \simeq \mathbb{Z}_2$. Here's where smoothness simplifies things: it is relatively easy to see the isomorphism between Čech cohomology and singular cohomology in the case of a smooth embedding, due to the tubular neighbourhood theorem. However, it requires some work to see that this is true also when $A$ is an embedded topological manifold. For details, see Bredon's Geometry and Topology's appendix "Euclidean Neighborhood Retracts". (Note that the isomorphism is not true in general, but it is true for reasonable spaces such as CW-complexes.)
