The procedure is fine.
I am assuming the system you are trying to solve is $\dot{x} = f(x)$, where $x\in \mathbb{R}^n$, and $f$ is smooth.
Let $C_K = \overline{B(0,K)}$. This is compact for all $k$, so choose a Lipschitz constant $L_K$ that would be valid on $C_{K+1}$. Choose $x_0 \in C_K$ which is the initial condition at some time $t_0 \in \mathbb{R}$. Then by the existence & uniqueness theorem for ODEs (see, eg, Kantorovich & Akilov, "Functional analysis", XVI, Section 4, Theorem 1) there exists $\epsilon>0$ such that a unique solution starting from $x_0$ exists on $[t_0,t_0+\epsilon]$. Furthermore, this $\epsilon$ is uniformly valid for all $x_0 \in C_K$, hence if $x(t_0+\epsilon) \in C_K$, the procedure can be repeated to extend the solution to $[t_0,t_0+2\epsilon],...,[t_0,t_0+(k+1)\epsilon]$, etc, as long as $x(t_0+k\epsilon) \in C_K$.
Let $T_K = \sup \{ t_1 | x(\xi) \in C_K, \ \forall \xi \in [t_0,t_1] \}$. $T_K$ is a non-decreasing sequence, so let $T_\infty=\lim_{K\to \infty} T_K$. If $T_\infty = \infty$, then a solution exists for all $t\geq t_0$, if not, then the solution becomes unbounded in finite time, ie, $\lim_{t\to T_\infty} \|x(t)\| = \infty$ (since $\|x(T_K)\|=K$).
A simple example of this unbounded behaviour in $\mathbb{R}$ is $\dot{x} = x^2$ with a positive initial condition.
Global uniqueness follows from local uniqueness. If $x,y$ are two solutions starting from $x_0$, then let $t^* = \sup \{ t\geq t_0 | x(t) = y(t) \}$. If $t^* < \infty$, then by considering the system starting from $x(t^*)$ (which is the same as $y(t^*)$), we see that there is a unique solution on some interval $[t^*, t^*+\delta]$, which contradicts the definition of $t^*$.
So we see that with a smooth field, you can always extend the solution in space, but not necessarily in time.