Significance of Diagonalization in Degenerate perturbation Theory

Yes and yes. For this particular problem, it is true that any method of finding eigenvalues will yield the answer you're looking for–though finding the eigenvalues is virtually synonymous with diagonalizing the matrix (the diagonalized matrix is simply the eigenvalues down the diagonal).

There is tremendous physical significance to the diagonalization of Hamiltonians. In condensed matter physics, for example, to diagonalize the Hamiltonian of a model is to understand its energy spectrum (gapped/gapless, degeneracies etc.), as well as its fundamental excitations, which is a lot of information. If you know the energy spectrum of a many-body system, you can conclude almost anything you would want to about its time evolution and thermodynamic properties.

Furthermore, subspaces of Hilbert space are defined by their shared quantum numbers (or equivalently, their symmetries). These subspaces show up in your Hamiltonian matrix as blocks, and in a finite-dimensional Hamiltonian they can be neatly visualized, and even manually constructed as a computational shortcut in the process of diagonalizing the full Hamiltonian.

In the degenerate perturbation problem you've cited, you are interested in diagonalizing the perturbed Hamiltonian in any degenerate subspaces. These are blocks of the perturbed Hamiltonian in the same matrix location as any blocks of the unperturbed Hamiltonian with a repeated eigenvalue. As we said earlier, the states in these block share any relevant symmetries. The perturbation can break one or more of these symmetries, resulting in energy corrections which lift the degeneracy. This fractures the degenerate subspace into several smaller, less degenerate subspaces. If any symmetry remains unbroken, these smaller subspaces may still themselves be degenerate.