Essentially the same as Marcos Cossarini's proof, but for readers unaware of 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} Then as usual, $S' - \lambda vv^T$ with $v=(x+\epsilon y)/\|x+\epsilon y\|$ has $rank(S')\le rank(S)-1$ and proceed by induction on the rank.