I am more new to math proofs and linear algebra.
I am trying to prove that for a symmetric positive block matrix such as:
$S=\begin{bmatrix}A&B\\ B^T&D\end{bmatrix}$
that for any eigenvalues $\lambda_i$ of the principal submatrices (A or D) are bounded by the smallest and largest eigenvalues of $S$ meaning $\lambda_S(min)\leq\lambda_i\leq\lambda_S(max)$.
I've looked through the proof of the Cauchy Interlace Theorem and that is way beyond my level of linear algebra knowledge and since what I am looking to prove is much more general, I was hoping I could be guided towards a more appropriate approach for both this proof and my understanding.
Your result (which, as I state in my comment, is less general than the Cauchy interlacing theorem) is fairly easy to show. Note that in general, for a symmetric matrix $M$: $$ \lambda_\max(M) = \max_{\|x\| = 1} x^TMx, \quad \lambda_\min(M) = \min_{\|x\| = 1} x^TMx. $$ With that established we could prove the result as follows. Let $m\times m$ be the shape of $A$ and $n\times n$ the shape of $D$. Let $U \subset \Bbb R^{m +n}$ denote the set of all vectors $x = (x_1,\dots,x_{m+n})$ for which $x_{m+1} = \cdots = x_{n} = 0$. In order to show that all eigenvalues of $A$ are less than the maximal eigenvalue of $S$, it suffices to argue that $$ \lambda_{\max}(S) = \max_{\|x\| = 1,\ x \in \Bbb R^{m +n}}x^TMx \geq \max_{\|x\|=1,\ x \in S} x^TMx = \max_{\|y\| = 1,\ y \in \Bbb R^m}y^TAy = \lambda_\max(A). $$ The proof that the eiegnvalues of $A$ are greater than the smallest eigenvalue of $S$ is similar, as are the proofs regarding the eigenvalues of $D$.