You cannot prove that $TU$ is self-adjoint because this is not true in general. You should try some other way to prove that all eigenvalues of $TU$ are real. For example, let us abuse the notations so that $T,U$ are also their respective matrix representation under the canonical basis.
- As $U$ is positive definite, it can be orthogonally diagonalized as $U=QDQ^T$ for some positive diagonal matrix $D$ and some real orthogonal matrix $Q$.
- Hence you may take a self-adjoint square root of $U$. That is, set $U^{1/2}=QD^{1/2}Q^T$, where $D^{1/2}$ is the entrywise square root of $D$.
- Show that $TU$ is similar to $U^{1/2}TU^{1/2}$ and argue that the latter has real eigenvalues.
An extension of the definition of "positive definite" to non-symmetric (real) matrices is to require $v^T M v \gt 0$ for all suitable nonzero vectors $v$. However see the discussion on this older Question about the lack of a conventionally adopted definition.
For real matrices this allows for non-self-adjoint (non-symmetric) examples. However the analogous "extension" for complex matrices, $v^* M v \gt 0$ for all nonzero vectors, turns out to imply $M$ self-adjoint (Hermitian), so it gives "nothing new".
As far as diagonalizability goes, neither requiring all eigenvalues positive nor requiring positive definiteness in the sense given above is sufficient to imply similarity to a diagonal matrix. Of course similar to a diagonal matrix would be implied by a full set of distinct eigenvalues (as each eigenvalue has at least one eigenvector), and it is equivalent to a basis of eigenvectors (since the matrix representation wrt to such basis is diagonal).
Here's an example of a nonsymmetric real matrix that (a) has only the positive eigenvalue 1 and (b) satisfies the "coercivity" condition $v^T M v \gt 0$ for all nonzero (real) vectors but (c) cannot be diagonalized:
$$ M = \begin{pmatrix} 1 & 0.1 \\ 0 & 1 \end{pmatrix} $$
Note that if $v^T = (x \; y)$, then $v^T M v = x^2 + 0.1xy + y^2$, which is easily shown to be a positive definite quadratic form, and thus the second (stronger) notion of a positive definite (but nonsymmetric) real matrix holds.
Best Answer
First note that eigenvalues are invariant under cyclic permutations, i.e, $\sigma(AB) = \sigma(BA)$ where $\sigma(X)$ denotes the set of eigenvalues of $X$. Then all we need to do is note $$ \sigma(AB) = \sigma(A^{1/2}A^{1/2} B) = \sigma(A^{1/2} B A^{1/2}) $$ but $\sigma(A^{1/2} B A^{1/2}) \in \mathbb{R}$ as $A^{1/2} B A^{1/2}$ is self-adjoint.