How can one know every Cauchy sequence in a complete metric space converges? I am new to Cauchy sequences. I stumbled onto them in the process of learning what a Hilbert space is.  As I understand it, a Cauchy sequence is a sequence whose elements become arbitrarily close to each other as the sequence progresses. 

But consider the following statement: 

A metric space $X$ in which every Cauchy sequence converges to an element of $X$ is called complete. 

How can we know every Cauchy  sequence converges?
$\Bbb R$ is a complicated example.  Let's instead consider a simple example.  Let $S$ be the space $\{7, 8\}$ with the usual distance.  (That is, $d(7,7) = d(8,8) = 0, $ and $d(7,8) = d(8,7) = 1$.)  I claim that $S$ is a complete space.

We are not going to show this by considering every cauchy sequence one at a time.  As you observed, that is impossible. 

Let $C$ be a cauchy sequence and the elements of $C$ are $c_0, c_1, \ldots$.  Since $C$ is a cauchy sequence, we have, for any positive $\epsilon$, that elements of the sequence that are far enough out will be closer than $\epsilon$.  Formally, for any given positive $\epsilon$ there is some integer $N$ so that for any $i,j&gt;N$, we have $d(c_i,  c_j) &lt; \epsilon$.

Take, for example, $\epsilon = \frac12$.  Then because $C$ is cauchy, we know there must exist some $N$ such that for any $i,j&gt;N$, we have $d(c_i , c_j) &lt; \frac12$, because that is what it means for $C$ to be cauchy; that is the definition.

But if we have  $d(c_i , c_j) &lt; \frac12$ then $d(c_i , c_j) = 0$, because the only  possible values of $d(\ldots)$ are $0$ and $1$, and it can't be $1$ because $1$ is not less than $\frac12$.  So it must be $0$.

So we conclude that if $C$ is cauchy, then for all sufficiently large
$i$, and $j$ we have $d(c_i, c_j) = 0$ and therefore $c_i = c_j$ for all sufficiently large $i$ and $j$. Thus, every cauchy sequence in $S$ is eventually constant.  It might have a mixture of $7$s and $8$s at the beginning, but there must come some point (that's $N$) after which all the elements are $7$s or all the elements are $8$s.  If this doesn't happen, then $C$ was not a cauchy sequence.

Since $C$ is eventually constant, there is some value, either $7$ or $8$, call it $L$, such that all sufficiently late elements of $c$ are equal to $L$.  That is, there is some $N$ such that for any $i&gt;N$, $c_i = L$, where $L$ is either 7, or 8, depending on which $C$ we actually chose.

Is is then trivial to show that $C$ converges to the value $L$. To show this we need to show that for any positive $\epsilon$, the terms of the sequence are eventually within $\epsilon$ of $L$.  In this case we can do even better; we know that the terms of $C$ are eventually equal to $L$.  $C$  is therefore a convergent sequence.  

Therefore, all cauchy sequences in $S$ converge, and $S$ is therefore a complete metric space.

The same argument goes through just as well if we take the more complicated space $\Bbb Z=\{\ldots, -1, 0, 1, 2, 3\ldots\}$ with the usual distance function $d(i,j) = |i-j|$; the proof is exactly the same.

Mathematics is not (usually) about considering a lot of examples.  It is about what properties logically imply which other properties.  Just as we can prove that every square number is positive without considering each square individually, we can also prove that every cauchy sequence of elements of $\Bbb Z$ must converge, even without considering each cauchy sequence individually.
You have your statement wrong: 

A metric space $X$ in which every Cauchy sequence that has a limit converges to an element of $X$ is called complete.

And you sense of what "Cauchy" means is a bit vague. For example, in the space $\Bbb{R}$ (the real line), the sequence $H_n \equiv \sum_{i=1}^n \frac{1}{i}$ has arbitrarily small distances between successive members, but is not is Cauchy because there for any given $m$ there is an $H_m$ with $m&gt;n$ that is arbitrarily far from $H_m$.  So this sequence it is not considered in deciding whether $\Bbb{R}$ is a complete space.

One way to prove that all Cauchy sequences that have a limit converge to an element of $X$ is to show that $X$ is compact, for that is a property of a compact space.  This is easy to see:  If $X$ is compact, then it contains all its limit points, and the limit of a Cauchy sequence is a limit point.  

It turns out that compactness is a bit of a stronger condition than is needed for completeness. For example, $\Bbb{R}$ is not compact. But since it does contain open sets including each of the points it contains, it is easy to see that $\Bbb{R}$ contains all its limit points, and therefore the limit of a Cauchy sequence of elements of $\Bbb{R}$ is in $\Bbb{R}$ so $\Bbb{R}$ must be complete.

I will admit that since you generally can't just test every possible Cauchy sequence, there are some spaces for which it is tough to prove whether they are complete or not. Sometimes very clever arguments are needed.