Here is a simple explanation not necessarily from linear algebra. We have
where $\|\cdot\|$ is simple euclidean norm. This is a constrained optimisation problem with Lagrange function:
here I took squares which do not change anything, but makes the following step easier.
Taking derivative with respect to $x$ and equating it to zero we get
the solution for this problem is the eigenvector of $A^2$. Since $A^2$ is symmetric, all its eigenvalues are real. So $x'A^2x$ will achieve maximum on set $\|x\|^2=1$ with maximal eigenvalue of $A^2$. Now since $A$ is symmetric it admits representation
with $Q$ the orthogonal matrix and $\Lambda$ diagonal with eigenvalues in diagonals. For $A^2$ we get
so the eigenvalues of $A^2$ are squares of eigenvalues of $A$. The norm $\|A\|_2$ is the square root taken from maximum $x'A^2x$ on $x'x=1$, which will be the square root of maximal eigenvalue of $A^2$ which is the maximal absolute eigenvalue of $A$.