This theory proof about instability of a point of equilibrium is not understandable for me, any help? -This theory is irritating me, because I don't understand it's logic.
Theorem: If in some neighboorhood $\mathbb O (0)$, exists a continuous, differentiable function $V(X), V(0)=0,$ such that the derivative based on the system: $X'=F(X)$ satisfies the inequality: $$\forall X \in \mathbb O(0): V'(X) \geq W(X)$$ where $W$ is some continuous, positive function on $\mathbb O (0),$ then the point of equilibrium $X=0$ is unstable.
PROOF:
In accordance with the conditions of the thereom, for every trajectory $X=X(t)$ of the system $X'=F(X)$ the function $V(X(t))$ is rising on the parts of the trajectory that are within $\mathbb K_r= \{X : \|X\|<r\}\subset \mathbb O(0).$Since $V$ is continuous on $\mathbb K_r,V(0)=0,\exists M, \mathcal E_0 > 0, \mathcal E \leq r:$ $$\forall X \in \mathbb K_{\mathcal E_0}:V(X)<M$$ 
Let $\delta \in (0, \mathcal E_0)$ be arbitrarily small. Choosing $X_0 \neq 0, \|X_0\|\leq \delta, V(X_0)>0$ and constructing a trajectory: $X=\varphi(t), \varphi(t_0)=X_0.$ Let's show that there exists $T>t_0$ such that $\|\varphi(T)\|> \mathcal E_0.$ Suppose the condradictory: $$\forall t >t_0 :\|\varphi(T)\ \leq \mathcal E_0.$$
Then, because of the rise of $V(\varphi(t))$**(From here on I  start having troubles)**
$$\forall t >t_0: V(\varphi(t))> V(\varphi(t_0))=V(X_0)>0$$ or 
$$\exists \beta \in (0, \mathcal E_0) \forall t \geq t_0: \|\varphi(t)\| \geq \beta$$ therefore 
$$V'(\varphi(t))\geq W(\varphi(t))\geq  \min_{\beta \leq \|X\|\leq \mathcal E_0}W(X)= \gamma > 0.$$ (The following line kills me!)
-Therefore function V is unlimited on $\mathbb K_{\varepsilon_0}$ a contradiction has arrised and we conclude that: 

$$\exists \varepsilon_0 > 0 \forall \delta >0, \exists X_0 \neq 0, \|X_0\| \leq \delta, \exists T > t_0, \|\phi(T)\|>\varepsilon_0$$ which is the definition of instability.

I would absolutely appreciate help on this and will give 100 rep to the first right answer, garanteed.
 A: Let's show that there exists $t>t_0$ such that $\|\varphi(t)\|> \mathcal E_0$. Let us suppose it is not the case and let us get a contradiction. 
So, suppose that: $$\forall t >t_0 :\|\varphi(t)\| \leq \mathcal E_0.$$
However as long as $\varphi$ remains in $\mathbb B_{\mathcal{E}_0}= \{X : \|X\|\leq\mathcal{E}_0\}$ we have that 
$$V'(\varphi(t)) \geq \gamma > 0$$
(see remark below for a proof).
Since V' is the derivative of V along the solutions of $X'=F(X)$, we have that the function $f$ defined by $f(t)= V(\varphi(t))$. has derivative  $f'(t)= V'(\varphi(t))$. 
So we have that $\forall t >t_0$, $\varphi$ remains in $\mathbb B_{\mathcal{E}_0}$ and so $f'(t) \geq \gamma > 0$. It is easy to prove that $$\lim_{t \to +\infty}f(t)=+\infty$$
So $\forall t >t_0$, $\varphi$ remains in $\mathbb B_{\mathcal{E}_0}$ and 
$$\lim_{t \to +\infty}V(\varphi(t))=+\infty$$
In particular, $V$ is not unbounded in $\mathbb B_{\mathcal{E}_0}$, which is a contradiction since $V$ is continuous and $\mathbb B_{\mathcal{E}_0}$ is compact.
Remark: We can prove that there is $\gamma > 0$ such that $\forall X\in\mathbb B_{\mathcal{E}_0}$, $V'(X) \geq \gamma > 0$ in the following way: 
We know that $\forall X \in \mathbb O(0): V'(X) \geq W(X)$ where $W$ is some continuous, positive function on $\mathbb O (0)$. And we have that $\mathbb B_{\mathcal{E}_0} \subseteq \mathbb O (0)$. So $W$ is a continuous, positive function on $B_{\mathcal{E}_0}$. So we have $\forall X\in\mathbb B_{\mathcal{E}_0}$
$$V'(X)\geq W(X)\geq\min_{Y\in B_{\mathcal{E}_0}}W(Y)> 0$$
Just make $\gamma=\min_{Y\in B_{\mathcal{E}_0}}W(Y)$.
