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.

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 $U_n$).
