The Cayley-Hamilton theorem by density

Step 1: matrices with distinct eigenvalues. Suppose $A$ has $n$ distinct eigenvalues $\lambda_1,\dots,\lambda_n\in\mathbb{C}$. Then $A=PDP^{-1}$ with $D=\operatorname{diag}(\lambda_i)$ and $P\in GL_n(\mathbb{C})$. For every polynomial $q$, $q(A)=Pq(D)P^{-1}$. Taking $q=\chi_A$,
$$\chi_A(A)=P\operatorname{diag}\bigl(\chi_A(\lambda_1),\dots,\chi_A(\lambda_n)\bigr)P^{-1}=0,$$
since each $\lambda_i$ is a root of $\chi_A$.

Step 2: continuity. The map $\Phi:A\mapsto\chi_A(A)$ from $M_n(\mathbb{R})$ to $M_n(\mathbb{R})$ is continuous: if $\chi_A(X)=X^n+c_{n-1}(A)X^{n-1}+\dots+c_0(A)$, each coefficient $c_j(A)$ is a polynomial in the entries of $A$, and $A\mapsto A^j$ is polynomial. So $\Phi$ is polynomial in the entries.

Step 3: density. The set $\Omega$ of matrices with $n$ distinct complex eigenvalues is dense (it is $\{\operatorname{Disc}\chi_A\ne0\}$, a nonempty polynomial open set). The continuous map $\Phi$ vanishes on the dense set $\Omega$, so $\Phi\equiv0$: $\chi_A(A)=0$ for all $A$.

Remark. The same argument proves the theorem over $\mathbb{C}$. Note that Step 1 genuinely needs complex diagonalisability, even for real $A$. Lesson. For polynomial identities in matrix entries, it suffices to check them on diagonalisable matrices.