Proving that a scheme is $\epsilon$-perfectly secret
There is no mistake.
Just prefix your work with the observation that any scheme is secure for $\epsilon\ge 1/2$ because all probabilities are less than or equal to 1.
Then consider the case $0&lt;\epsilon&lt;1/2$. Repeat your argument, then note that $2(1-\epsilon)&gt;1$ and so $\epsilon/2(1-\epsilon)&lt;\epsilon$.