Suppose $\epsilon$ > 0 and $\delta$ = min {1,$\frac{\epsilon}{10}$}
Is it true that if 0 < |x -1 |< $\delta$, then this implies 0 < |x -2 |< $\epsilon$?
I am saying yes and here is why. We are given that 0 < |x -1 |< $\delta$, so -1 < |x -2 |< $\frac{\epsilon}{10}$-1, which is
-1< |x -2 |< $\frac{\epsilon}{10}$-1
Now $\frac{\epsilon}{10}$-1 is always less than $\epsilon$. And that is why I say it is true.
Your argument is wrong - here's a counterexample. Take $\epsilon={1\over 2}$, so $\delta={1\over 20}$. Let $x=1+{1\over 100}$. Then $0<\vert x-1\vert<\delta$, but $\vert x-2\vert={99\over 100}>\epsilon$.
Specifically, your error is that your "... so ..." is not justified at all.