Chain of Implications for Continuity and Boundedness Consider the following definitions:

>
  1). Somewhere Locally Bounded: 

$\exists p \in X, \exists \epsilon >0, \exists \delta >0, \forall q \in X: d(p,q)< \delta \Rightarrow d(f(p),f(q)) < \epsilon$.

>
  2). Locally Bounded: 

$\forall p \in X, \exists \epsilon_p < \infty, \exists \delta_p >0, \forall q \in X : d(p,q)<\delta \Rightarrow d(f(p),f(q)) < \epsilon$. 

>
  3). Continuous:

$\forall p \in X, \exists \epsilon >0 \exists \delta_{p,\epsilon} >0, \forall q \in X :
d(p,q)<\delta \Rightarrow d(f(p),f(q)) < \epsilon$.

>
  4). Uniformly Continuous:

$\forall \epsilon >0, \exists \delta_\epsilon >0, \forall p \in X, \forall q \in X:
d(p,q)< \delta \Rightarrow d(f(p),f(q)) < \epsilon$.

>
  5). Lipschitz Continuous:

$\exists C>0, \forall \epsilon >0 \forall p \in X, \forall q \in X:
d(p,q)< \frac{\epsilon}{C} \Rightarrow d(f(p),f(q))< \epsilon$. 
Alternatively, we can also write 
$\forall \delta>0, \forall p,q \in X, d(p,q)<\delta \Rightarrow d(f(p),f(q))< C \cdot \delta$.
What I am asked to do is to find 4 examples of $(X,Y,f)$ showing $(5) \nLeftarrow (4) \nLeftarrow (3) \nLeftarrow (2) \nLeftarrow (1)$. 
Here is my attempt at the complete solution:
$(5) \nLeftarrow (4)$.
Not all uniformly continuous functions are Lipschitz. 
Let us consider a continuous function on a compact set which has a vertical asymptote. Let $f(x)=\sqrt[3]{x}$ on $[-1,1]$. Since $[-1,1]$ is compact, we have uniform continuity. But for $x_n= \frac{-1}{n^3}$ and $y_n = \frac{1}{n^3}$, we have 
$\displaystyle {|\frac{f(x_n)-f(y_n)}{x_n-y_n}| = |\frac{\frac{-2}{n}}{\frac{-2}{n^3}}| = n^2}$. However there is no $C$ such that $n^2 \leq C$ for all $n$, so the function $f$ cannot be Lipschitz continuous. 
$(4) \nLeftarrow (3)$. 
Continuity does not imply Uniform Continuity.
Let $f(x)=x^2$. This function is continuous, but not uniformly over $I=(0,\infty)$.
$\forall x_0 \in I, \forall \epsilon >0, \exists \delta >0, \forall x \in I$, such that 
$|x-x_0|< \delta \Rightarrow |x^2-x_0^2|< \epsilon$.
Let us fix $x_0$ and let $a=x_0+1$ and $\delta = \min(1,\frac{\epsilon}{2a})$.
Choose $x\in I$. Assuming $|x-x_0|< \delta$, we have $|x-x_0|<1, \quad x<x_0+1=a.$ So we get 
$|x^2-x_0^2|= |(\frac{1}{\delta}+\frac{\delta}{2})^2-(\frac{1}{\delta})^2| = |\frac{1}{\delta^2} +1+ \frac{\delta^2}{4} - \frac{1}{\delta^2}| = 1+ \frac{\delta^2}{4} > 1 = \epsilon$ as required. 
$(3) \nLeftarrow (2)$
Locally Bounded does not imply Continuity.
Let $f(x)= \frac{x^2-4}{x-2}$. Then we have that $f(x)$ is locally bounded, say by (1,3), but $f(x)$ is not continuous at 2.
$(2) \nLeftarrow (1)$
For the last implication, I do not know how to start.
This is what I have so far, please feel free to correct any technical errors. I apologize for the very long post, but I have been struggling with this problem for a while now and would like to finally solve it.  
As always, any assistance is greatly appreciated. Thanks in advance. 
 A: You could try to translate those definitions into a more 'geometrical' form. 
$1)$ tells you that there is some fixed point $p$, where the for all $q$ close enough to $p$ the values $f(p),f(q)$ are not too far apart.
$2)$ tells you that for every point $p$, for some $q$ close enough to $p$ the values $f(p),f(q)$ are not too far apart.
So, a way to go about this is to notice that the first one is a purely local property - given a function which is constant on a certain interval and completely arbitrary on the rest, can still fulfill the first condition, whereas the second property is more global - at least, you have the local property everywhere on the domain. 
I'd suggest looking at any not locally bounded function (you might want to think about rationals here, anything continuous will not work), and then switch it up with a locally bounded one somewhere - thus, you will get your contradiction.
Also, please note that your counterexample to $3)\Leftarrow 2)$ does not hold - the function $\frac{x^2-4}{x-2}=\frac{(x+2)(x-2)}{x-2}=x+2$ is perfectly continuous. Again, I'd recommend separating rational numbers and irrational numbers, and try to find a non-continuous, bounded function. 
A: For your $3)\not\Leftarrow 2)$ example, try the Dirichlet Function:
$$
D(x)=\left\{\begin{array}{lc}
1 & :x\in\mathbb{Q} \\ 0 & :x\not\in\mathbb{Q}
\end{array}\right.
$$
It tends to be great for continuity counterexamples, as it is discontinuous everywhere (and can be easily modified to be continuous, and even differentiable, at a finite or countable number of points). It should be obvious that is is bounded, as $|D(a)-D(b)|\leq 1$ for any $a,b\in\mathbb{R}$.
As for showing that $2)\not\Leftarrow 1)$, you only need to find a function that is not locally bounded at a single point. The easiest example is probably
$$f(x)=\left\{\begin{array}{lc}
0 & : x=0 \\
\frac{1}{x} & :x\neq0
\end{array}\right.$$
at $x=0$.
A: You can fix the counterexample involving $f(x)=\frac{x^2-4}{x-2}$ by specifying that $f(2) = \frac{5}{2}$.
Here is a function that is somewhere locally bounded but not locally bounded:
$$
f(x) = \left\{
\begin{array}{cc}
x^q & x \mbox{ rational }, x= p/q \mbox{ with gcd}(p,q) = 1 \\
0 & x \mbox{ irrational }
\end{array}
\right.
$$
$f(x)$ is locally bounded at $x=0$ (or indeed at any point $x=p$ with $0 \leq p < 1$ thus it is somewhere locally bounded.  But $f(x)$ is not locally bounded:  Consider point $p$ at  $x=2$, where for points arbitrarily close to $p$, $f(x)$ is close to  $2^n$ for arbitrarily large $n$. 
