How can I see that the rank of $\mathrm{SU}(N)$ is $N-1$? Is there a way to understand intuitively why this must be the case? Or is there a kind of easy-to-see proof of this result? Thank you in advance.

It seems easier to work on the level of the Lie algebras, the tangent to a maximal torus being precisely a maximal abelian subalgebra. The Lie algebra of $SU(n)$ is $n\times n$ traceless skew-Hermitian matrices. If they commute, they are simultaneously diagonalizable (by a version of spectral theorem), so in some basis are diagonal. That means (recalling the traceless condition) that the dimension is at most $n-1$. Of course it is then exactly $n-1$.
