Why $\mathbb{RP}^2$ can not be embedded to $\mathbb{R}^3$? Is there any answer of this question around basic theory of differentiable manifolds? 
 A: The simplest solution is via Alexander duality, which shows immediately that every surface in $S^3$, in particular in $\mathbb{R}^3$, is orientable.
A: Every closed smooth hypersurface   $X\subset \mathbb R^n$ (=submanifold of dimension $n-1$), compact or not, has an equation,  i.e.  $X=f^{-1}(0)$ for some  smooth $f:\mathbb R^n\to \mathbb R$ satisfying  $df(x)\neq0$ for all $x\in X$.
In particular $X$ is orientable so that  $\mathbb P^2(\mathbb R) $, which is not orientable, cannot be embedded into $\mathbb R^3$   
(Non-) bibliography
The fact that every closed hypersurface  in $\mathbb R^n$ has an equation is unfortunately not proved nor even mentioned in any book I know on differential manifolds or differential geometry.
It is however rather elementary and depends on methods developed  60 years ago in complex analysis : I wrote a proof here. (See also there, where only orientability of hypersurfaces is proved.)
A: The Stiefel-Whitney class of $\Bbb RP^2$ is $(1+a)^3=a^2+a+1 \in H^*(\Bbb RP^2,\Bbb Z/2)$ where $a$ is the unique nonzero element in $H^1$. The inverse of this element in $H^*(\Bbb RP^2,\Bbb Z/2)$ is $1+a$. As the top Stiefel-Whitney class of the normal bundle of an embedding into $\Bbb R^n$ vanishes (c.f. Milnor-Stasheff Cor. 11.4) $1+a$ is not the Stiefel-Whitney class of the normal bundle to  an embedding into $\Bbb R^3$, hence there is no embedding $\Bbb RP^2\rightarrow \Bbb R^3$. The theory of characteristic classes gives numerous obstructions to embeddings, immersions, cobordisms and other similar geometric constructions. A place to start is the Milnor-Stasheff book Characteristic Classes cited in this answer. 
The above argument easily generalizes to show $\Bbb RP^{2^n}$ does not embed in $\Bbb R^{2^{n+1}-1}$, hence the inequality in the strong Whitney embedding theorem is sharp for dimensions arbitrarily large.
