Help understanding a proof So the textbook I'm reading just stated three definitions of continuity:
a) f is continuous at a (using neighborhoods)
b) the epsilon-delta definition
c) If $x_n$ is a any sequence of elements of $D(f)$ which converges to a, then the sequence $(f(x_n))$ converges to $f(a)$
Then it proceeded to show that the three statements are equivalent. I'm having trouble with understanding the part of the proof where a) implies c).
The portion of the proof goes as follows:
We shall argue that if condition a) does not hold, then condition c) does not hold. If a) fails, then there exists a neighborhood $V_0$ of $f(a)$ such that for any neighborhood $U$ of $a$, there is an element $x_U$ belonging to $D(f)\cap U$ but such that $f(x_U)$ does not belong to $V_0$. For each natural number consider the neighborhood $U_n$ of $a$ defined by $U_n=\{x\in \mathbb{R^p}:\|x-a\|<\frac{1}{n}\}$;from the preceding sentence, for each $n$ in $\mathbb{N}$ there is an element $x_n$ belong to $D(f)\cap U$ but such that $f(x_n)$ does not belong to $v_0$. The sequence $(X_n)$ just constructed belongs to $D(f)$ and converges to a, yet none of the elements of the sequence $(f(x_n))$ belong to the neighborhood $V_0$ of $f(a)$. Hence we have constructed a sequence for which the condition c) does not hold. This shows that part c) implies a).$\square$
How do we know that the neighborhood $U_n$ allows for an element $x_n$ to belong to $D(f)\cap U$ but that $f(x_n)$  is not in the neighborhood $V_0$? Also, where did the sequence $(x_n)$ get constructed? Any help is appreciated.
 A: Firstly, note that this is a proof that (c) implies (a), and not the converse. 
You are assuming that (a) fails, so there is at least one neighborhood of $ f (a) $ whose inverse image does not contain any open neighborhoods of $ a $.  Since this neighborhood will be fixed for the rest of the argument, we'll call it $ V_0$.  
Since $ f^{-1}[V_0] $ does not contain any open neighborhood of $ a $, it follows that every open neighborhood of $ a$ contains at least one point whose image under $ f $ does not lie in $ V_0$.  In particular, each $ U_n $ must contain at least one point $ x_n $ satisfying 
$$ f (x_n)\notin V_0. $$
Since we always have $ x_n\in U_n $, it is obvious that $\lim_{n\to\infty} x_n=a $.  But since each $ x_n$ was chosen so that $f (x_n) $ is not in $ V_0$, it follows that the sequence $ \langle f (x_n) \vert n\in \mathbb{N} \rangle $ cannot possibly converge to $ f (a) $ (or any point in $ V_0$ for that matter).  
In short, if (a) fails, then (c) must also fail. Thus, by contrapositive reasoning, (c) implies (a).

In terms of motivation, it helps to bear in mind some canonical examples. For example, consider the step function $f:\mathbb{R}\to\mathbb{R}$ given by $$f(x)=\left\{\begin{array}{ll} 1& \textrm{ if } x\geq 0 \\ 0 & \textrm{ otherwise.}\end{array}\right.$$ In this case $a=0$ and $f(a)=1$. 
The fact that (a) fails means that there is a neighborhood of $1$ whose inverse image is not open. In this case, the inverse image of $(1/2,\infty)$ is the interval $[0,\infty)$. Every neighborhood of $0$ contains at least one point $x$ satisfying $f(x)\leq 1/2$.
What would failure of (c) look like?  It would be a sequence of numbers converging to $0$ whose images do not converge to $1$.  So that's pretty clear - for any sequence of negative numbers, the image of that sequence is the constant sequence $0$. So if we can find a sequence of negative numbers which converges to $0$, then we will have shown that (c) fails.  
Our problem is to use the failure of (a) to construct such a sequence. It's pretty clear that the failure of (a) allows us to find negative numbers close to $0$ (since the points satisfying $f(x)\leq 1/2$ are precisely the negative numbers). So to get a sequence of such numbers that converges to $0$, we just need a sequence of neighborhoods of $0$ which get close to $0$.  The sequence $$U_{n}=\big\{x\in\mathbb{R}\, \big\vert\, \Vert x\Vert < 1/n\big\}$$is one of many sequence of neighborhoods that will make this work.
A: Since (a) fails, there exists some fixed neighborhood $V_0$ of $f(a)$ such that for any neighborhood $U$ of $a$, there exists an $x_U \in U \cap D(f)$ such that $f(x_U)$ is not in $V_0$.
The sequence $(x_n)$ is constructed using that fact. For each $U_n$, there exists some $x_n \in U_n$ satisfying the right properties (remember, we can find an $x_U$ for any neighborhood $U$; in particular, we can find an $x_n$ for each $U_n$).
