If $f$ is bounded then solutions of ODE are defined for all time? Let's say we have the following Cauchy Problem: 
$x' = f(t,x)$ 
$x(t_0) = x_0$
Is this statement true?
If $\exists$ $K$ such that $||f(t,x)|| \leq K$ for all $(t,x) \in R \times R^n$ $\implies $ The solution is defined for all time
 A: If $f$ is continuous then by the fundamental theorem of calculus we write the DE as  an integral equation: $$x(t)=x(t_0)+\displaystyle\int_{t_0}^tf(s, x(s))ds.\quad (1)$$
By the Cauchy-Peano existence theorem the IVP has a solution defined in some open interval $(t_0-\delta, t_0+\delta)$ for some $\delta>0$. By continuation of solutions we can arrive a (maximal) solution with the maximal interval of existence $(\alpha, \beta)$. If we show that $\alpha=-\infty$ and $\beta=\infty$ we say that the solution is defined for all $t$. Let us prove that $\beta=\infty$. (The case $\alpha=-\infty$ is similar and left to you). The proof relies on contradiction. Suppose that $\beta$ is finite, i. e., $\beta<\infty$. Since $|f|\leq M$, It follows from (1) that $$|x(t_2)-x(t_1)|\leq M|t_2-t_1|,\quad (2)$$
where $\alpha<t_1<t_2<\beta$. Now if we pick a Cauchy sequence $(t_n)$ such that $t_n<\beta$ for each $n$ converging $\beta$, from (2) we see that $x(t_n)$ is also a Cauchy sequence. It is not difficult to see that $\lim_{t\to\beta}x(t)$ exists and it $b$. Let $y$ be a solution to the IVP:
$x' = f(t,x)$ 
$y(\beta) = b$.
Then applying the Cauchy-Peano existence theorem once again we see that this problem has a solution in some interval $(\beta-\delta_1, \beta+\delta_1)$ for some $\delta_1$. This means that $y(t)$ is an extension of $x(t)$ defined in $(\beta-\delta_1, \beta+\delta_1)$, so $x(t)$ is not maximal which is a contradiction. By a similar approach you can show that $\alpha=-\infty$.
