Essentially the same as Marcos Cossarini's proof, but without Cauchy-Schwarz.

   Let $S\in R^{n\times n}$ symmetric of rank $\ge 1$.
By compactness let $x\in R^n$ with $\|x\|=1$ be such that
    $\|Sx \| = \max_{u\in R^n: \|u\|=1} \|Su\|$
 and set $y = Sx/\|Sx\|$.
    Set $\epsilon=+1$ if $y^Tx \ge 0$ and $\epsilon=-1$ if $y^Tx<0$. Then
$\|x+\epsilon y\|^2 \ge 2$ so that $x+\epsilon y \ne 0$. With $\lambda=\epsilon \|Ax\|$ we have
    $Sx = \epsilon \lambda y$ and
    \begin{align}
  \|S(x+\epsilon y) - \lambda(x+\epsilon y)\|^2
&=\|\epsilon Sy - \lambda x\|^2
\\&= \|S y\|^2 + \lambda^2 - 2 \epsilon \lambda y^TS x
\\&= \|Sy\|^2 + \lambda^2  - 2\lambda^2
\\&\le 0 \text{ since } \|Ay\|\le \|Ax\|=|\lambda| \text{ by definition of }x.
    \end{align}