Is Lipschitz condition necessary for existence and uniqueness of a solution for a first order differential equation? Is the Lipschitz condition only a sufficient condition for existence and uniqueness of $\frac{dy}{dx}=f(x,y), y(0)=y_0$? Is it also necessary?

No, it is NOT necessary - It is only sufficient.
For example, if $f$ is JUST continuous and $f(x)>0$, for all $x\in\mathbb R$, then 
$$
x'=f(x), \quad x(\tau)=\xi, \tag{1}
$$
enjoys GLOBAL uniqueness, for all $(\tau,\xi)\in\mathbb R^2$.
Say $\varphi,\psi : I \to\mathbb R$, satisfy $(1)$, where $I$ is an open interval and $\tau\in I$. Then
$$
1=\frac{\varphi'(t)}{f\big(\varphi(t)\big)}=\frac{\psi'(t)}{f\big(\psi(t)\big)},
\quad \text{for all $t\in I$}.
$$
Set 
$$
F(x)=\int_\xi^x\frac{dx}{f(x)}, \quad x\in\mathbb R.
$$
Clearly, $F$ is continuously differentiable and strictly increasing (hence 1-1), in $\mathbb R$, and
$$
F\big(\varphi(t)\big)=\int_\tau^t\frac{\varphi'(t)}{f\big(\varphi(t)\big)}=\int_\tau^t\frac{\psi'(t)}{f\big(\psi(t)\big)}=F\big(\psi(t)\big), \quad\text{for all $t\in I$}.
$$
Thus, as $F$ is 1-1, $\varphi(t)=\psi(t)$, for all $t\in I$.
So, $(1)$ enjoys global uniqueness, while $f$ does not satisfy Lipschitz condition! 
