Is the space $C[0,1]$ complete? In order to prove $C[0,1]$ is complete, my functional analysis book says: 

It is only necessary to show that every Cauchy sequence in $C[0,1]$ has a limit.

It goes on by supposing $\{x_n\}$ is a Cauchy sequence in $C[0,1]$. Then for each fixed $t \in [0,1]$ $|x_n(t)-x_m(t)| \le \|x_n - x_m\| \to 0$, so $\{x_n(t)\}$ is a Cauchy sequence of real numbers. I guess this inequality makes sense, since difference of function sequences would be larger than pointwise difference (at a specific $t$) of those functions.. But I did not understand how the author could conclude the sequence  $\{x_n(t)\}$ is Cauchy. 
Second question: the explanation continues, 
"Since the set of real numbers is complete, there is a real number $x(t)$ to which the squence converges; $x_n(t) \to x(t)$". 
I know set of real numbers is complete, but how can $\{x_n(t)\}$ represent the entire set of real numbers?
 A: This is simpler to prove if we split the claim up in parts. I'm not giving proofs because others here have already done that.
Theorem 1a. A uniformly Cauchy sequence of $\mathbb K$-valued bounded functions is pointwise convergent to a bounded function.
Theorem 1b. A uniformly Cauchy sequence of $\mathbb K$-valued bounded functions is uniformly convergent to its pointwise limit.
Corollary 2. $B(X \to \mathbb K)$ is a Banach space when equipped with the uniform norm.
Theorem 3. Let $X$ be a Hausdorff space, $(Y,d)$ a metric space and $f$ and $f_n$ functions from $X$ to $Y$. If $(f_n)$ converges uniformly to $f$ on $X$ (*) and if each $f_n$ is continuous at $x \in X$, then $f$ is continuous at $x$.
Corollary 4. $C_b(X \to \mathbb K)$, the set of bounded continuous functions on a Hausdorff space $X$, is a closed subspace of $B(X \to \mathbb K)$, so it too is a Banach space.
(*): i.e. for all $\varepsilon > 0$ there exists an $N \in \mathbb N$ such that $x \in X \land \mathbb N \owns n \geq N \implies d(f_n(x),f(x)) < \varepsilon$.
A: This is a particular case of the more general statement that if $K$ is a compact metric space then $C(K)$ with the topology induced by the $\sup$-norm  is complete.
To prove this you do the following:
You need to show that the limit of every sequence that is Cauchy with respect to $\| \cdot \|_\infty$ is in $C(K)$. So let $f_n$ be a Cauchy sequence in $C(K)$. First you need to find out what it might converge to. In this case (and in many other cases) the limit function that it converges to in norm is the same as its pointwise limit function. So:
(i) Prove that the pointwise limit function $f$ exists. What does that mean? It means that $f(k)$ is finite for all points $k$ in $K$. So let $k$ be a fixed point in $K$. Then consider $f_n(k)$. Note that this is a sequence of real numbers. Let $\varepsilon > 0$ and now remember that you picked $f_n$ to be a Cauchy sequence with respect to $\| \cdot \|_\infty$. So you have $$ | f_n(k) - f_m(k) | \leq \sup_{k \in K} | f_n(k) - f_m(k) | = \| f_n - f_m \|_\infty < \varepsilon$$
Which shows that $f_n(k)$ is a Cauchy sequence in $\mathbb{R}$ with respect to the standard metric. But you know that $\mathbb{R}$ is complete so you know that the limit of $f_n(k)$ is in $\mathbb{R}$. This means that the pointwise limit $f(k) = \lim_{n \to \infty} f_n(k)$ exists.
(ii) Next you go on to showing that $f_n$ also converges to $f$ in norm, that is, $\| f_n - f \|_\infty \xrightarrow{n \to \infty} 0$.
In other words: for every $\varepsilon > 0$ you need to find an $N$ such that for $n > N$ you have $\| f_n - f \|_\infty \leq \varepsilon$.
For this let $\varepsilon > 0$ and then fix an $N$ such that for $n,m \geq N$ you have $\| f_n - f_m \| < \frac{\varepsilon}{2}$. You can do this because $f_n$ is Cauchy by assumption. Then use the triangle inequality to get
$$ \| f_n - f \|_\infty \leq \| f_n - f_N \|_\infty + \| f - f_N \|_\infty$$
By how you picked $N$ you have $\| f_n - f_N \|_\infty < \frac{\varepsilon}{2}$.
You also know that for $m \geq N$ you have $\| f_m - f_N \|_\infty < \frac{\varepsilon}{2}$. From this you can conclude that $\lim_{m \to \infty} \| f_m - f_N \|_\infty \leq \frac{\varepsilon}{2}$.
But this means you have shown that $$ \| f_n - f \|_\infty \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} =\varepsilon$$
for $n > N$, which means exactly that $\| f_n - f \|_\infty \xrightarrow{n \to \infty} 0$.
(iii) In order to conclude that $C(K)$ is complete, the last thing you need to prove is that $f$ is continuous and therefore actually in $C(K)$. So let $\varepsilon > 0$ and $k$ in $K$. Now you want to find $\delta > 0$ such that for $x$ with $d(k,x) < \delta$ you have $|f(k) - f(x)| < \varepsilon$.
Use the triangle inequality to get 
$$ |f(k) - f(x)| \leq |f(k) - f_n(k)| + |f_n(k) - f_n(x)| + |f_n(x) - f(x)|$$
for some $n$. Now you pick $n$ such that $|f(k) - f_n(k)| < \frac{\varepsilon}{3}$ and $|f_n(x) - f(x)| < \frac{\varepsilon}{3}$. Then the $\delta$ you are looking for is now going to be the $\delta$ such that $|f_n(k) - f_n(x)| <\frac{\varepsilon}{3}$ which you know exists because $f_n$ are continuous. Hence
$$ |f(k) - f(x)| < \varepsilon$$ and therefore $f$ is continuous. 
Note that this works for $K$ compact because then the continuous functions on $K$ are bounded and so $\|\cdot\|_\infty$ is actually a norm. 
Using a very similar argument you can show that the space of bounded continuous functions on an arbitrary topological space $X$ is complete. I did this in another post here.
Hope this helps.
A: The norm we are working with is $\newcommand{\abs}[1]{|#1|}\newcommand{\norm}[1]{\|#1\|}\norm f= \sup_{x\in[0,1]}\abs{f(x)}$. (You didn't specify the norm in your question, but I guess it's very improbable that you would work with some different norm.)

Then for each fixed $t \in [0,1]$ $|x_n(t)-x_m(t)| \le ||x_n - x_m|| \to 0$, so $\{x_n(t)\}$ is a Cauchy sequence of real numbers. I guess this inequality makes sense, since difference of function sequences would be larger than pointwise difference (at a specific $t$) of those functions.. But I did not understand how the author could conclude the sequence  $\{x_n(t)\}$ is Cauchy.

Well, you have just shown that $|x_n(t)-x_m(t)|\to 0$. This is precisely the definition of Cauchy sequence.
The reason why the inequality $\abs{x_n(t)-x_m(t)}\le\norm{x_n-x_m}$ is the definition of $\norm{\cdot}$. For every bounded function $f$ you have $\abs{f(t)} \le \sup_{x\in[0,1]} \abs{f(x)}=\norm f$.

I know set of real numbers is complete, but how can $\{x_n(t)\}$ represent the entire set of real numbers? 

You do not need anything like that. Now you have fixed some $t\in[0,1]$ and you're working with one real sequence $\{x_n(t)\}$. This is a sequence of real numbers, by completeness of $\mathbb R$, such sequence converges if it is Cauchy.
After we have shown that the sequence $\{x_n(t)\}$ is convergent, we can denote the limit (for this fixed value of $t$) by $x(t)$.

Perhaps it might be useful for you to have a look at quite detailed proof of a similar results in Matt's answer here.
A: Fix $t_0$, $\epsilon > 0$. Since $x_n$ are Cauchy, there exists $N > 0 | \forall n,m > N$ we have $$||x_n - x_m || := \sup_t |x_n(t) - x_m(t)| < \epsilon$$In particular, for $t_0$, and the same $N$, for all $n,m > N$ we have $|x_n(t_0) - x_m(t_0)| < \sup_t |x_n(t) - x_m(t)| < \epsilon$. 
Now we know that $x_n(t_0)$, just a sequence of real numbers obtained by evaluating the functions $x_1, x_2, x_3 \ldots$ at $t_0$ is Cauchy. As a Cauchy sequence of real numbers, it must be a convergent sequence. 
