There are obviously many ways to prove your statement. Some of the comments suggest to use the following:
Let $v$ be an eigenvector to the eigenvalue $\lambda$. Set $U=\mathbb R v$ and write $\mathbb R^n=U\bot U^\bot$. Then $A(U)\subseteq U$ and $A(U^\bot)\subseteq U^\bot$ (you use symmetry here). Therefore we can restrict $A$ to $U^\bot$, get a symmetric matrix and proceed by induction. This is somewhat the standard proof of the spectral theorem.
Another way - I think a way closer to what you asked - would be the following lemma:
If $x\in Ker(\lambda I-A)^k$, then $x\in Ker(\lambda I-A)$.
For simplicity I will proof the case $k=2$. Let $x\in Ker(\lambda I-A)^2$ and $y=(\lambda I-A)x$. We want to show $y=0$. We have
$$\lambda \left<y,x\right >=\left<y,Ax+y\right >=\left<y,Ax\right >+\left<y,y\right >=\left<Ay,x\right >+\left<y,y\right >=\lambda\left<y,x\right >+\left<y,y\right >$$
Which means $y=0$.
If you believe that there is a basis of generalised eigenvectors you are done.