A theorem in topology involving sequences I have two questions.
1) How to prove the following theorem.
2) Could someone give me an example when one would use this and what for?
Theorem 4.22. Let $(X, d)$ be a metric space, let $x_n,\space n ≥ 1$, be a sequence in
$X$ and let $x$ be a point in $X$. The sequence $x_n$ converges to $x$ if and only if, for every $\epsilon > 0$, there is a positive integer $N$ such that $d(x_n, x) < \epsilon $, for all
$n ≥ N$.
Requested Edit:
Def'n
1) A sequence in a topological space X is a function from
$\Bbb N = {1, 2, 3, . . .} $ to X. Usually, we write $x_n,\space n ≥ 1$, or $x_1, x_2, . . .$ for a sequence instead of $f(n)$.
2) If $x_n,\space n ≥ 1$, is a sequence in a topological space X and x
is a point of X, we say that the sequence has limit x and write
$\lim_{n\to \infty} x_n = x$.
if for every open set U containing x, there exists a positive integer N such
that $x_n$ is in $U$, for every $n ≥ N$. We also say that the sequence converges
to x.
3) Let $(X, d)$ be a metric space. The collection of all sets of the
form $B(x, r)$, where x is in X and $r > 0$ is a basis for a topology on X. This
is called the metric topology.
 A: Note that $$d(x_n,x)< \epsilon \iff x_n\in B(x,\epsilon)$$
Also $U$ is an open neighborhood of $x$ $\iff$ $U$ contains an open ball $B(x,\epsilon)$  
A: Ok, so the proof of the theorem goes like this.
Assume that $(x_n)$ converges toward $x$, it means that by definition, for any open set $U$ containing $x$, there exists $N$ such that for $n\geq N$ $x_n\in U$. In particular, if I fix $\epsilon>0$, I can apply this to $U:=B(x,\epsilon)$ which leads to the conclusion. 
Assume now that for any $\epsilon>0$ there exists $N_{\epsilon}$ such that for any $n\geq N_{\epsilon}$ $d(x_n,x)<\epsilon$. Now if you take any open set $U$ containing $x$, I claim that by definition of an open set in a metric topology, you will be able to find $\epsilon_0>0$ such that $B(x,\epsilon_0)$ is contained in $U$. In particular, for any $n\geq N_{\epsilon_0}$, $x_n\in U$. This is true for any open set $U$ containing $x$, thus $(x_n)$ converges.
For your second question, this theorem has only a theoretical importance. You have, a priori, two notions of convergence on a metric space, a convergence coming from the topology (the one that you defined with open sets) and the one coming from the metric (the one with the distance itself). This theorem tells you that they define exactly the same limits and same converging sequences. 
A: So the key is to note that, for a metric space, you not only have open sets like in any normal topology, but you have open sets that can be thought of as these open "disks" of radius e. 
It's a nice added advantage of working with metric spaces since you get that much more information in what your open sets (at the basis layer) look like. 
In reality, its not that every open sets is a perfect "disk/circle" -- except at the basis layer. The rest of the open sets are either a "disk/circle" or created from a union/intersections of a collection of "disks/circles". 
Hence, as noted in the answer above, if U is an open neighborhood of x then U must contain some disk of radius e around x that fits inside U. (since U can be built from some collection of disks via intersections/unions) 
