Picard Theorem for Locally Lipschitz functions In order to apply the local Picard theorem in this question, one of the conditions I must satisfy is that $|f|\leq M$. 

Which part of the question can this be inferred from?
According to the Hint, you need to show that there exists $h &gt; 0$, such that the problem
$$
x' = f(x), \quad x(b) = p_0 \qquad \qquad (1)
$$
has a unique solution on the interval $(b-h, b+h)$.

Since $p_0 \in W$ and $W$ is open, there exists sufficiently small neighbourhood $U$ of the point $p_0 = x(b)$, such that $U \subset W$. 

Since $f$ is continuously differentiable on $W$ and $U$ is a "strict" subset of $W$, there exists $M &gt; 0$ such that $|f(x)| &lt; M$ for all $x \in U$. Moreover, $f$ is, of course, Lipschitz in $U$.

Let us denote $D := \mathbb{R} \times U$. Then we can directly apply this theorem (all assumptions will be satisfied) and obtain some $h &gt; 0$, such that (1) possesses a unique solution $x(t)$ on the interval $(b-h, b+h)$. This is the desired result.

That is why you don't need the boundedness of $f$ in the whole $W$ (and, in general, you surely don't have it).
According to the Hint, you need to show that there exists $h &gt; 0$, such that the problem
$$
x' = f(x), \quad x(b) = p_0 \qquad \qquad (1)
$$
has a unique solution on the interval $(b-h, b+h)$.

Since $p_0 \in W$ and $W$ is open, there exists sufficiently small neighbourhood $U$ of the point $p_0 = x(b)$, such that $U \subset W$. 

Since $f$ is continuously differentiable on $W$ and $U$ is a "strict" subset of $W$, there exists $M &gt; 0$ such that $|f(x)| &lt; M$ for all $x \in U$. Moreover, $f$ is, of course, Lipschitz in $U$.

Let us denote $D := \mathbb{R} \times U$. Then we can directly apply this theorem (all assumptions will be satisfied) and obtain some $h &gt; 0$, such that (1) possesses a unique solution $x(t)$ on the interval $(b-h, b+h)$. This is the desired result.

That is why you don't need the boundedness of $f$ in the whole $W$ (and, in general, you surely don't have it).