If $A$ is symmetric, then $I+\epsilon A \succ 0$ if $\epsilon$ is sufficiently small
Hint: You are taking a very intricate approach to the problem. This could be made to work, but I would suggest instead that work along the following lines.
- Note/show that there exists a constant $C>0$ such that for all $x \neq 0$, $\left|\frac{x^TAx}{x^Tx}\right| \leq C$.
- Note that $x^T(I + \epsilon A)x = x^Tx + \epsilon \cdot x^TAx$ and select $\epsilon$ accordingly.