Show that the function $f(x) = \frac{1 }x$ is uniformly continuous on the interval $[1, 2]$ using only the definition.
My Approach: For all $\epsilon > 0$ there exits $\delta > 0$ such that if $|x − y| < \delta$ then $|f(x) − f(y)| < \epsilon$.
So $|\frac{1}x - \frac{1}y| = |\frac{y-x}{xy}| < \delta|\frac{1}{xy}|$
Since $x,y \in [1,2]$, then we may say $\delta|\frac{1}{xy}| < \delta$ (since the highest $\frac{1}{xy}$ can be is then $\frac{1}{1*1}$.
So set $\epsilon= \delta$, and the proof is done. I hope :)
The idea is fine, but you need to adjust a few things. The point of an $\varepsilon$-$\delta$ proof is that you need to tell me, given some $\varepsilon>0$, who the $\delta$ is. Saying "there exists $\delta>0$" can never be a proof on itself, unless you are using some previous result.
What you have shown is that it is enough to take $\delta=\varepsilon$.