You cannot diagonalize real symmetric 2x2 matrices in a continuous way.

Let us restrict to the set $Sym^r(R)$ of 2x2 symmetric matrices with two distinct real eigenvalues. Then we have the following result.

Theorem There is no continuous function $f:Sym^r(R) \rightarrow SO_2(R)$ such that for all $M\in Sym^r(R)$, $f(M)Mf(M)^{-1}$ is diagonal.

Proof Let us restrict our attention to the $SO_2$-conjuguacy class $O_A$ of some fixed diagonal matrix $A\in Sym^r$. Let $D$ the set of diagonal matrices. If there were such a f, then the map $(A,D)\rightarrow f(A)D$ would give a homeomorphism between $O_A\times (D\cap SO_2)\rightarrow SO_2$. But $D\cap SO_2$ is not connected (it contains only $id$ and $-id$).

The complex case with hermitian matrices uses simple connectedness (you end up with $S^2\times S^1 \simeq S^3$). And the result holds in all dimension. This may be the occasion to speak about connectedness of matrix groups.