Proving that $\mathbb R^3$ cannot be made into a real division algebra (and that extending complex multiplication would not work)

Assume that $D$ is a 3-dimensional division algebra over $\Bbb{R}$. Let $a\in D\setminus\Bbb{R}$. Consider the linear mapping $\rho_a:z\mapsto az, z\in D,$ from $D$ to itself. Let $M$ be the matrix representing $\rho_a$ (with respect to some basis). The eigenvalue polynomial $$ \chi_a(x)=\det(xI_3-M)\in\Bbb{R}[x] $$ is monic of degree three. Thus $\lim_{x\to\pm\infty}\chi_a(x)=\pm\infty$. Therefore, by continuity of $\chi_a(x)$, there exists a real number $r$ such that $\chi_a(r)=0$. This means that the mapping $$ L:D\to D, z\mapsto az-rz=(a-r)z $$ has a non-trivial kernel. Therefore the element $a-r$ cannot be invertible. Because $a\notin\Bbb{R}$ we have $a-r\neq0_D$. This contradicts the assumption that $D$ is a division algebra.

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Edit: The above was a bit of overkill for the task at hand (=proving that complex multiplication cannot be extended to a 3D-space). A simpler argument follows.

If the multiplication of $D$ is an extension of the multiplication of $\Bbb{C}$, then $D$ has a subalgebra isomorphic to $\Bbb{C}$. Therefore $D$ has a structure of a (left) vector space over $\Bbb{C}$. Thus $D$ is a finite dimensional vector space over $\Bbb{C}$. But this implies that the dimension of $D$ as a vector space over $\Bbb{R}$ is an even number.

I will also add a more "topological" proof but requires some familiarity with manifolds:

For a multiplication to "expand" the complex multiplication we would require that $|u_1\times u_2|=|u_1||u_2|$ which is saying that $S^2$ would take the structure of a group (as is the case in $\mathbb{C}$ and $S^1$). But since $S^n$ are manifolds , this would $S^2$ a Lie group and we know that for every Lie Group $TM=M\times M$ , namely is is parallelizable.

Using the (very deep) fact that only $S^1,S^3,S^7$ are parallelizable we get that the only multiplications like $\mathbb{C}$ exist in $\mathbb{R^2},\mathbb{R^4},\mathbb{R^8}$ which ofcourse correspond to the complex numbers, quartenions and octonions.

Notice that for tha case of $S^2$ and in general $S^{2k}$ we can just use the hairy-ball theorem to show that they aren't parallelizable!