Proving that $f: E\to \mathbb R$, defined by $f(x)=\frac{1}{x-x_0}$ is not uniformly continuous on $E$. Let $E\subset \mathbb R$ and $x_0\notin E$ is a limit point of $E$. 
It is to be proven that $f: E\to \mathbb R$, defined by $f(x)=\frac{1}{x-x_0}$ is not uniformly continuous on $E$.
I tried to prove it as follows: 
We fix an $\epsilon \gt 0$. Since $x_0$ is a limit point of $E$, for any $\delta\gt 0, \exists y\in E$ such that $|y-x_0|\lt \delta/4$ and $ |y-x_o|\lt \frac{1}{\epsilon+\frac 4{\delta}+|\epsilon-\frac 4{\delta}|} $ 
and
$\exists x\in E$ such that $|x-x_0|\lt \frac{3\delta}4$ and $|x-x_0|\gt \frac{1}{\frac 12(\epsilon+\frac 4\delta)+\frac 12|\epsilon-\frac 4\delta|}\tag{1}$ 
It follows that $|f(y)|\gt \epsilon+\frac 4{\delta}+|\epsilon-\frac 4{\delta}|$ and $|f(x)|\lt \frac 12(\epsilon+\frac 4{\delta})+\frac 12|\epsilon-\frac 4{\delta}|$
It follows that: $|x-y|\le |x-x_0|+|y-y_0|\lt 3\delta/4+\delta/4=\delta$ and that $|f(y)|-|f(x)|\gt \frac 12(\epsilon+\frac 4{\delta})+\frac 12|\epsilon-\frac 4{\delta}|=\max\{\epsilon, \frac 4\delta\}\gt \epsilon\implies |f(y)-f(x)|\gt \epsilon$.
Since $\delta\gt 0$ is arbitrary, it follows that $f$ is not uniformly continuous on $E$. 
Is my proof correct? Thanks. 
Edit:
In $(1)$: $x$ has  been chosen in such a way that it remains in $(x_0-3\delta/4, x_0+3\delta/4)$, considering that $\max\{\epsilon,\frac 4\delta\}=\frac 12 (\epsilon +\frac 4\delta)+\frac 12|\epsilon -\frac 4\delta|\gt \frac 4\delta\implies \frac 1{\max\{\epsilon,\frac 4\delta\}}\lt \frac \delta 4$.
Now there are two cases: 
1) $|x-x_0|\le \frac {\delta }4$ 
In this case, we have: $\frac 1{\max\{\epsilon,\frac 4\delta\}}\lt |x-x_0|\le \frac{ \delta}4\lt \frac {3\delta }4$ 
2) $|x-x_0|\gt \frac{\delta }4$. 
In this case, we have $\frac \delta 4\lt |x-x_0|$ and $\frac 1{\max\{\epsilon,\frac 4\delta\}}\lt |x-x_0|\lt \frac{3\delta }4$. 
That is, in both the cases we have $\frac 1{\max\{\epsilon,\frac 4\delta\}}\lt |x-x_0|\lt \frac{3\delta }4$ and this validates our choice of $x$ in $(1)$.
 A: Let's try to solve the more general problem.
Suppose that $E\subset \mathbb{R},\ x_0\notin E$ and $x_0$ is a limit point of $E$. Suppose we are given the function $f:E\to \mathbb{R}$ with the following property: in any deleted neighborhood $f$ is unbounded, i.e. for any $r>0$ the function $f(t)$ is unbounded on $E\cap ((x_0-r,x_0+r)\setminus \{x_0\})$.
One can check that your function $E\mapsto \mathbb{R}$ defined by $t\mapsto \dfrac{1}{t-x_0}$ has this property.
Take any $\delta>0$ consider the neighborhood $W:=E\cap ((x_0-\delta/2,x_0+\delta/2)\setminus \{x_0\})$. Pick a point $x_2\in W$. Since $f$ is unbounded on $W$ then one can find $x_1\in W$ such that $|f(x_1)|>1+|f(x_2)|$.
Therefore, for $\forall \delta>0$ we found $x_1,x_2\in E$ with $|x_1-x_2|<\delta$ but $|f(x_2)-f(x_1)|\geq |f(x_1)|-|f(x_2)|>1.$
Hence $f$ is NOT uniformly continuous on $E$.
$\textit{Remark:}$ The condition that $x_0$ is a limit point of $E$ should be used when we want to prove that $f$ is unbounded in any nbhd of $x_0$.
A: Since $x_0\not\in E$ is a limit point of $E$, for any $\epsilon\gt0$, there is an $x_\epsilon\in E$ so that $|x_\epsilon-x_0|\lt\epsilon$.
Now if $f(x)=\frac1{x-x_0}$ were uniformly continuous on $E$, we should be able to find a $\delta\gt0$ so that if $|x-y|\lt\delta$, then $|f(x)-f(y)|\lt1$.
However, for any $\delta\gt0$, by the definition of $x_\epsilon$,
$$
r=|x_{\delta/2}-x_0|\lt\delta/2\tag1
$$
By $(1)$,
$$
|f(x_\delta)|=\frac1r\tag2
$$
Consider the point $x_{r/(r+1)}$. By the definition of $x_\epsilon$ and $(1)$,
$$
\left|x_{r/(r+1)}-x_0\right|\lt\frac{r}{r+1}\lt r\lt\delta/2\tag3
$$
and so
$$
\left|f\!\left(x_{r/(r+1)}\right)\right|\gt1+\frac1r\tag4
$$
By the Triangle inequality and $(1)$ and $(3)$,
$$
\begin{align}
\left|x_{\delta/2}-x_{r/(r+1)}\right|
&\le\overbrace{|x_{\delta/2}-x_0|\vphantom{\left|x_{r/(r+1)}\right|}}^{\lt\delta/2}+\overbrace{\left|x_{r/(r+1)}-x_0\right|}^{\lt\delta/2}\tag{5a}\\[6pt]
&\lt\delta\tag{5b}
\end{align}
$$
However, by the Triangle inequality and $(2)$ and $(4)$,
$$
\begin{align}
\left|f\!\left(x_{r/(r+1)}\right)-f(x_{\delta/2})\right|
&\gt\overbrace{\left|f\!\left(x_{r/(r+1)}\right)\right|}^{\gt1+\frac1r}-\overbrace{\left|f(x_{\delta/2})\right|\vphantom{\left(x_{r/(r+1)}\right)}}^{\frac1r}\tag{6a}\\[6pt]
&\gt1\tag{6b}
\end{align}
$$
Since $\delta\gt0$ was arbitrary, $(5)$ and $(6)$ show that $f$ is not uniformly continuous.
A: I do not understand your proof, but I think it is easier to just use sequences. Since $x_0$ is a limit point of $E$, there is a sequence $(x_n)$ such that $x_n\to x_0.$ It is no loss of generality to assume $|x_n-x_0|<1$ for all integers  $n$. Then, with $\epsilon=1,$  if $\delta>0$ is given, there is an integer $N$ such that $|x_N-x_0|<\min\{1,\delta\}$ and then of course $\frac{1}{|x_N-x_0|}>1,$ which contradicts the definition of uniform continuity.
