Let $f$ be a continuous Lipschitz function, then exists unique solution for $x'=f(t,x)$ with $x(t_0)=x_0$ for all $t\in \mathbb{R}$

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Let $f:\mathbb{R}\times\mathbb{R}^n\rightarrow \mathbb{R}^n$ be continuous and Lipschitz. How do I prove that there exists unique solution of the IVP $x'=f(t,x),\quad x(t_0)=x_0$

for all $t\in \mathbb{R}$.

I tried to use the Banach fixed-point theorem, but I think that $M=C(\mathbb{R},\mathbb{R}^n)$ is not a complete metric space whit the supremum norm. The reason I tried this is to define $T:M\rightarrow M$ as $T(z)(t)=x_0+\int_{t_0}^{t} f(s,z(s))\,ds$ and prove that there exist a fixed point using the following inequality

$$\|T(x)(t)-T(y)(t)\|\leq \frac{L^n \lvert t-t_0\rvert^n}{n!}d(x,y),$$

where $L$ is the Lipschitz constant for $f$ and $d(x,y)=\sup\limits_{t\in \mathbb{R}}\|x(t)-y(t)\|$.

Can somebody give me a hand?