Understand Picard-Lindelöf Proof I am trying to understand the Picard-Lindelöf from my book which uses the fixed point theorem.
The task is trying to find $x \in C(a,b)$ in open interval $(a, b)$ containing $t_0$ such that it satisfies the integral equation $x(t) = x_0 + \int_{t_0}^{t}f(s, x(s))ds$ for all $t \in (a, b).$ 
Here $C(a,b)$ denotes the set of bounded continuous functions and for $a < t_0 < b$ and $\beta > 0$ define $K = [a, b] \times \overline{B(x_0, \beta)}$
Define $X \subset C(a,b)$ be set of continuous functions $x: (a,b) \rightarrow R^n$ such that $\|x(t)-x_0\| \leq \beta$ for $t \in (a,b).$ 
The task is to show that X is complete so for a Cauchy sequence $\{x_k\} \in X$ since $C[a, b]$ is complete, it converges to a function $x \in C(a,b)$ and the goal remains to show that $x \in X$. 
Then $\|x(t)-x_0\| = lim_{p \rightarrow \infty}\|x_p(t)-x_0\| \leq \beta$ for $t \in (a,b)$ since $\|x_{p}(t) - x_0\| \leq \beta$ for $t \in (a,b).$ Thus this shows that $x \in X$. I understand everything up to this point.
What I don't understand is the next part:
Moreover there is no loss of generality in looking for fixed points in X and not in C(a,b). Indeed if $x: (a,b) \rightarrow R^n$ is continuous and satisfies the integral equation, then $\|x(t)-x_0\| \leq \|\int_{t_0}^{t}f(s,x(s))ds\| \leq|t-t_0|M \leq (b-a)M$ where $M = max\{\|f(t,x)\|: (t,x) \in K\} < \infty$ because f is continuous and K is compact. This shows that if $x \in C(a,b)$ then it belongs to X for some $\beta$.
What I don't understand is why the second argument is necessary. Didn't the first argument already show that if $x(t) \in C(a,b) \implies x(t) \in X$?

The big picture: 


*

*A certain subset $X\subset C[a,b]$ is introduced; we'll be looking for a solution of the integral equation in $X$.

*$X$ is shown to be closed in $C[a,b]$. Hence, $X$ is a complete metric space with the induced metric.

*The restriction of the integral operator to $X$ is shown to be a contraction of $X$ into  $X$.  

*From 2 and 3 we conclude that the equation has a unique solution in $X$.

*We rule out the possibility of solutions in $C[a,b]\setminus X$. This part is   needed for the uniqueness part of the theorem. (We already know there is exactly one solution in $X$, but what if there are solutions outside of $X$?)


Part 5 is what you are asking about.
