If $f(x) \cdot x < 0$ for all $x \in \partial B_R(0)$, then the IVP $x' = f(x)$, $x(0) = x_0$ has a global solution. I have a homework problem that asks

If $f : \mathbb{R}^n \to \mathbb{R}^n$ is continuously differentiable and satisfies
  $$ f(x) \cdot x < 0 \quad \quad \text{for all } x \in \partial B_R(0) $$
  for some $R > 0$ (where $B_R(0)$ is the open ball centered at the origin of radius $R$), then the initial value problem
  $$ x' = f(x), \quad x(0) = x_0 $$
  has a global solution for every $x_0 \in B_R(0)$.

I know that $f$ is locally Lipschitz because $f$ is continuously differentiable and $B_R(0)$ is open and connected, so Picard's existence theorem says there is in fact a solution.  Say the maximal interval of existence is $I = (\alpha, \beta)$, and suppose for the sake of contradiction $\beta \neq \infty$.  I have a theorem at hand that says:
For any compact subset $K \subseteq B_R(0)$, there exists a $\beta_K < \beta$ such that $x(t) \in B_R(0) \setminus K$ for all $t \in (\beta_K, \beta)$.
Now I think I should consider the sequence of compact sets $\left\{ \overline{B_{R - \frac{1}{n}}(0)} \right\}_{n=1}$.  For each one there is some $\beta_n < \beta$ such that $R - \frac{1}{n} < \| x(t) \| < R$ for every $t \in (\beta_n, \beta)$...
This is where I am stuck.  The assumption about the dot product is almost surely going to come into play soon, but I don't see how to use it to show some contradiction.  I feel like I am very close.
 A: First of all, for every IVP enjoying uniqueness, there exists a solution defined in a maximum interval $[0,T)$, where $T\le\infty$. We need to show that $T=\infty$.
Next, observe that, due to continuity of $f$ and compactness of $\partial B_R(0)$, there exist an $\varepsilon>0$, such that
$$
x\cdot f(x)<0, \quad \text{for}\,\,\, R-\varepsilon\le \lvert x\rvert\le R.
$$
Next, shall show that $x(t)\in B_{R}(0)$, for all $t\in [0,T)$. Otherwise, there would be an $t_0$, such that
$x(t_0)\in\partial B_{R(0)}$ and $x(t)\in B_{R}(0)$, for $t<t_0$. In particular, there would exist a $\delta>0$, such that $t_0-\delta>0$ and
$$
R-\varepsilon\le\lvert x(t)\rvert< R, \quad\text{for}\,\,\, t\in [t_0-\delta,t_0).
$$
But in such case we would have that
$$
0<\frac{1}{2}\big(\lvert x(t_0)\rvert^2-\lvert x(t_0-\delta)\rvert^2\big)
=\int_{t_0-\delta}^{t_0}x(t)\cdot x'(t)\,dt
=\int_{t_0-\delta}^{t_0}f\big(x(t)\big)\cdot x(t)\,dt<0,
$$
which is a contradiction. Hence $x(t)\in B_{R}(0)$, for all $t\in [0,T)$.
Next, use the fact that: 
If the function $x :[0,T)\to\mathbb R^n$, with $T<\infty$ satisfies the equation $x'=f(x)$, and $x$ bounded, then the limit $\lim_{t\to T} x(t)$ exists in $\mathbb R^n$, and $x$ 
extends as a continuously differentiable function in $[0,T]$.
Say  $\lim_{t\to T} x(t)=\xi$.
Then, the IVP
$$
x'=f(x), \quad x(T)=\xi,
$$
enjoys existence is some interval $(T-a,T+a)$, and its solution extends the previous one, which had domain $[0,T)$, to $[0,T+a)$.
Thus $T$ can not be a finite number, and hence $x$ is definable on $[0,\infty)$.
A: Here is the idea: a solution defined on an interval $(\alpha, \beta)$ with $\beta < \infty$ 
cannot be extended past $\beta$ if and only if $\lim_{t \nearrow \beta} x(t) = \infty$, that is, the solution escapes at infinity as $t$ inches up to $\beta$. 
Assume that the solution escapes to infinity on the interval  $(\alpha, \beta)$
What you have with $S(0,R)$ is a compact barrier. You got yourself a function $b(x)= ||x||^2$ with a compact sublevel set $\{ b \le R^2\}$ so that $\nabla b(x)\cdot f(x) <0$ for all $x$ so that $b(x) = R^2$ ( the vector field $f$ points towards the lower values of $b$). It should be intuitively clear that if you start initially  in $\{ b \le R^2\}$ you can never escape it. Therefore, one cannot reach infinity in finite time, hence every solution starting in the ball $B(0,R)$ is globally defined.
I am looking at your solution and notice that you are not fully using the condition of escaping to infinity. At some point close to $\beta$ the solution will be on the sphere.
Take the first such point. There are such points. Why?
Don't just consider compacts inside the ball, any compact will work, since $f$ is globally defined. So at some moment before $\beta$ you are outside so you must have crossed the sphere earlier. Take the first moment of crossing.
Note that a solution cannot escape the ball $\bar B(0,R)$ even if the weaker condition 
$x \cdot f(x) \le 0$ holds.  More generally, a solution cannot escape the sublevet set $\{b(x) \le \rho\}$ if on the level set $b(x) = \rho$   we have $\nabla b \ne 0 $ and 
$\nabla b (x) \cdot  f(x) \le 0$.
