Showing a solution is unique 
Let $F:\;[a,b]\times \mathbb{R}^n\to\mathbb{R}^n$ continuous where $0\in [a,b]$ and for some $K>0$: 
  $$||F(t,x)-F(t,y)||\leq K||x-y||$$ for all $x,y$ and all $t\in [a,b]$.
I would like to show that there is unique continuous
  $f:[a,b]\to\mathbb{R}^n$ solving: $$f(t)=\int_{0}^t F(s,f(s))\; ds$$

Idea:
I know a weaker statement: if we have the conditions above but instead, for $B_R$ a closed ball of radius $R$ centered at $0:$
$$\mathbf{1.} \;F:\;[a,b]\times B_R\to\mathbb{R}^n\quad \mathbf{2.} \;\sup ||F||\leq R(b-a)^{-1}$$
If a continuous $f:[a,b]\to B_R$ solves the equation, it is unique.
Now I realise I cannot just "let $R\to\infty$", but is it possible to use the fact above to prove the statement?
 A: Proof of Uniqueness:
Assume that $f,g: [a,b]\to \mathbb R^n$ satisfy the integral equation:
$$
f(t)=\int_0^tF\big(s,f(s)\big)\,ds, \quad
g(t)=\int_0^tF\big(s,g(s)\big)\,ds.
$$
Clearly, $f(0)=g(0)=0$, and
$$
f(t)-g(t)=\int_0^t \Big(F\big(s,f(s)\big)-F\big(s,g(s)\big)\Big)\,ds, \quad t\in[a,b].
$$
We use the fact that $\Big\|\int_c^d G(t)\,dt\,\Big\|\le \int_c^d\|G(t)\|\,dt,$
and obtain that, for $t\ge 0$:
$$
\|f(t)-g(t)\|\le\int_0^t \Big\|F\big(s,f(s)\big)-F\big(s,g(s)\big)\Big\|\,ds\le \int_0^t K\|f(s)-g(s)\|\,ds \quad t\ge 0.\tag{1}
$$
Set $\varphi(t)=\int_0^t \|\,f(s)-g(s)\|\,ds$. Then $\varphi'(t)=\|\,f(t)-g(t)\|$,
and hence $\varphi$ satisfies the inequality:
$$
\varphi'(t)\le K\varphi(t), \quad t\ge 0.
$$
Note that $\varphi(0)=0$. Multiply the above by $\mathrm{e}^{-Kt}$ and obtain
$$
\mathrm{e}^{-Kt}\big(\varphi'(t)- K\varphi(t)\big)\le 0, \quad t\ge 0,
$$
or
$$
\big(\mathrm{e}^{-Kt}\varphi(t)\big)'\le 0, \quad t\ge 0,
$$
and integrating the continuous function above in $[0,t]$ we obtain
$$
\mathrm{e}^{-Kt}\varphi(t)-\varphi(0)=\int_0^t\big(\mathrm{e}^{-Ks}\varphi(s)\big)'\,ds\le 0, \quad t\ge 0.
$$
But $\varphi(0)=0$, and thus
$$
0=\varphi(t)=\int_0^t \|\,f(s)-g(s)\|\,ds, \quad t\ge 0,
$$
and differentiating
$$
0=\varphi'(t)= \|\,f(t)-g(t)\|, \quad t\ge 0.
$$
The case $t\le 0$, is treated similarly.
A: Note that your integral equation is equivalent to the differential equation
$$\frac {{\rm d} f} {{\rm d} t} = F(t, f(t))$$
with the initial condition $f(0) = 0$: if $f$ solves the differential equation, then integrate to get
$$\int \limits _0 ^t F(s, f(s)) {\rm d} s = \int \limits _0 ^t \frac {{\rm d} f} {{\rm d} t} (s) {\rm d} s = f(t) - f(0) = f(t) ;$$
conversely, if $f$ solves the integral equation, then it is derivable (with continuous derivative, because $F$ is continuous), so you may derive the integral equation to get the differential one written above. Also, $f(0) = \int \limits _0 ^0 F(s, f(s)) {\rm d} s = 0$.
It is now enough to use the Picard-Lindelöf to obtain what you want, many proofs of which are easy to find with a simple web search.
