I know a few examples of metric tensors (flat, spherical, hyperbolic, Schwartzchild, Kerr, Kerr-Newman) and there's a common property between them. All of them can be expressed as
\begin{equation} g_{\mu\nu}=diag(g_t,g_x,g_y,g_z) \end{equation}
In other words, if $\mu\ne\nu$:
\begin{equation} g_{\mu\nu}=0 \end{equation}
Can the metric tensor always be diagonalized? Is there a proof/disproof for this?
It is always possible to find coordinates such that the metric is diagonal at some chosen event (indeed, it can even be Minkowski at a chosen event). In more than 3 dimensions it is not always possible to find coordinates such that the metric is diagonal at all events.
The reason that a method of diagonalization at one event cannot extend to all events is that it places constraints on the way the coordinates have to vary, and they cannot always satisfy those constraints while also remaining smooth (i.e. such that metric is differentiable to all orders).
(By the way, I was surprised to see Kerr metric on your list: are you sure about that?)
Yes, the metric tensor can always be diagonalized. That is because the metric is by definition real symmetric, thus it can always be diagonalized by some appropriately chosen orthogonal matrix:
\begin{equation} g = Q g_{d} Q ^{-1} \end{equation}
where $g = g^{T}$ the original metric, $g_{d}$ the diagonalized form, and $Q$ the orthogonal matrix ($Q^{T} = Q^{-1}$).
This is a specific case of the spectral theorem for normal matrices, which says that a general matrix can be diagonalized by unitary matrices if and only if it is normal:
\begin{equation} M^{\dagger} M = M M^{\dagger} \iff M = U D U^{\dagger} \end{equation}
where $D$ is diagonal and $U^{\dagger} = U^{-1}$. Obviously in the case that $M \in GL(n, \mathbb{R})$, then $U$ is also a real matrix, and thus the unitarity condition becomes an orthogonality one. And if $M = M^{T}$, then it is trivially normal, thus it can always be diagonalized.
A metric tensor, a doubly-covariant tensor representing a quadratic form rather than a linear map, is diagonalized by $g\to {\rm diag}=M^T g M$ rather than by $g\to M^{-1} g M$. For example consider $$ (dx)^2 +6 dx\,dy + 13 (dy)^2= \left[\begin{matrix} dx & dy\end{matrix}\right] \left[\begin{matrix}1 & 3 \cr 3 &13\end{matrix}\right] \left[\begin{matrix}dx \cr dy\end{matrix}\right] $$ When we diagonalize a quadratic form we do not need to find eigenvalues and eigenvectors,we just compete squares: $$ ds^2 =(dx)^2 +6 dx\,dy + 13 (dy)^2= (dx+ 3 dy)^2+ 4(dy)^2. $$ so changing coordinates $$ d\xi=dx+3dy \\ d\eta= 4 dy $$ or $$ \left[\begin{matrix}d\xi \cr d\eta\end{matrix}\right]=\left[\begin{matrix}1 & 3 \cr 0 &4\end{matrix}\right] \left[\begin{matrix}dx \cr dy\end{matrix}\right] $$ the metric becomes $$ ds^2 = (d\xi)^2- (d\eta)^2= \left[\begin{matrix} d\xi & d\eta\end{matrix}\right] \left[\begin{matrix}1 & 0 \cr 0 &1\end{matrix}\right] \left[\begin{matrix}d\xi \cr d\eta \end{matrix}\right]. $$ and the matrix diagoinalization reads $$ \left[\begin{matrix}1 & 3 \cr 3 &13\end{matrix}\right] =\left[\begin{matrix}1 & 0 \cr 3 &4\end{matrix}\right] \left[\begin{matrix}1 & 0 \cr 0 &1\end{matrix}\right]\left[\begin{matrix}1 & 3 \cr 0 &4\end{matrix}\right]. $$
When the metric depends on location the transformation will depend on where you are.
Of course you can go to the effort of finding the eigenvalues and eigenvectors so that you end up with an orthogonal matrix for $M$, but the Kerr example by @KP99 shows that this is not necessary.
