Picard–Lindelöf theorem conditions not met with unique IVP solution

Question

According to Picard–Lindelöf theorem, for $E$ Banach space and $\Omega$ an open subset of $\mathbb R \times E$, the IVP $$\begin{cases} y^\prime(t) = f(t,y(t))\\ y(x_0)=y_0 \end{cases}$$

has a unique local solution providing that $f$ is uniformly Lipschitz continuous in $y$ and continuous in $t$.

Is there an example of $f(y)$ being continuous in $\mathbb R$, $\mathcal C^1$ in $\mathbb R \setminus \{0\}$, not differentiable in $0$ ($f^\prime(0)=\infty$), but such that the IVP $$\begin{cases} y^\prime(t) = f(y(t))\\ y(0)=a \end{cases}$$ has a unique solution?

Note: the existence of a solution is provided by Peano existence theorem.

Answer 1

Take $$ \begin{cases} y'(t) = 1 + \sqrt{\lvert y(t) \rvert} \\ y(0) = a, \end{cases} $$ where $a$ is arbitrary. The unique solution $[t \mapsto \varphi(t)]$ is given in the implicit form $$ t = \int\limits_{a}^{\varphi(t)} \frac{d\eta}{1 + \sqrt{\lvert \eta \rvert}}, $$ whereas $f(y) = 1 + \sqrt{\lvert y \rvert}$ is continuous on $\mathbb{R}$, $C^1$ on $\mathbb{R} \setminus \{0\}$, and not differentiable at $y = 0$.