Proof that the $\lim\limits_{x \to 2}\dfrac{1}{x} = \dfrac{1}{2}$ using the $\epsilon$-$\delta$ definition of limits (verification). Prove that the $\lim\limits_{x \to 2}\dfrac{1}{x} = \dfrac{1}{2}$ using the $\epsilon-\delta$ definition of limits.
$$
\\ \begin{align}
\\ &\textrm{Let } \forall \epsilon > 0
\\ &\textrm{Choose } \delta = \min{\{1, 2\epsilon \}}
\\ &\textrm{Assume } 0 < |x - 2| < \delta :
\\ \end{align}
$$
$$
\\ \begin{align}
\\ \left|\frac{1}{x} - \frac{1}{2}\right| &< \epsilon
\\ \frac{|2 - x|}{|2x|} &< \epsilon
\\ |-1(x - 2)| &< \epsilon|2x|
\\ |x - 2| &< \epsilon|2x|
\\ \end{align}
$$
$$
\\ \begin{align}
\\ |x - 2| &< 1
\\ -1 < x - 2 &< 1
\\ 1 < x &< 3
\\ \end{align}
$$
$$
\\ \begin{align}
\\ |x - 2| &< \epsilon|2(1)|
\\ |x - 2| &< 2\epsilon
\\ \end{align}
$$
$$
\\\therefore \delta \leq 2\epsilon
$$
Right, so I start by taking $|f(x)−L|<ϵ$. I then isolate $|x−2|$ to the left. I then limit $|x−2|$ to be less than one and then find a range of $x$ values which satisfy the inequality. Then I plug in the smallest $x$ value to minimise the value of $ϵ$. I make the conclusion that $δ≤2ϵ$. Am I excluding or misplacing steps? I'm fairly new to this whole thing.

Here you have to show that for each $ \epsilon >0$, there exists $ \delta >0 $ such that for each $ x\in Domn(\frac{1}{x}) $ if $ 0<|x-2|<\delta $ then $ \left | \frac{1}{x}-\frac{1}{2}\right |<\epsilon $.
So begin with arbitrary $ \epsilon >0 $. 
Notice that if $ 0<|x-2|<1 $ then $ 1<|x|<3 $ and hence $\frac{1}{3}<\frac{1}{|x|}<1$.
Now choose $ \delta =\min\{1,2\epsilon\} $. Then clearly $ \delta >0 $.
Now suppose $ 0<|x-2|<\delta $.
Then $ \left | \frac{1}{x}-\frac{1}{2}\right |=\frac{|x-2|}{2|x|}<\frac{|x-2|}{2}<\frac{2\epsilon}{2}=\epsilon $.
Therefore $$ \lim_{x\rightarrow 2}\frac{1}{x}=\frac{1}{2} .$$
