ODE Hyperbolic fixed point study Here is a problem I can not solve : 
Let $X(x, y)$ be a $C^{\infty}$ vector field defined on $\mathbb{R}^2$ such that $X(x, y) = (y + a(x, y), x+ b(x, y))$ where $a, b : \mathbb{R}^2 \to \mathbb{R}$ are bounded and  $a(0) = b(0) = 0$ and $da(0) = db(0) = 0$.
We assume there exist periodic points $ p_n = (x_n, y_n) \to (0, 0)$ (and $p_n \neq 0$) such that the solution of $ (x',y') =X(x,y)$ and $(x(0),y(0))=p_n$ is $T_n$ periodic. I want to show that $T_n \to \infty$
What I've done : 


*

*We have $\|(x',y')\| \le \| (x,y)\| + C $ with $C$ a constant so $\phi ^t$ is defined on $\mathbb{R}$

*$0$ is an hyperbolic point

*By the Hartman Grobman theorem I proved that there is a $\varepsilon$ such that the ball $B(0,\varepsilon)$ does not contain any periodic orbit. 
Argue by contradiction. Suppose there is an extraction $\phi$ such that $T_{\phi (n)}$ is bounded by $T_{\infty}$.
Then I named $L_n$ the length of the orbit $\phi ^t (p_n)$ and I want to show that if $n$ is large enough then $L_n\le \varepsilon $ :
$$ L_n= \int_0^{T_n} \sqrt{x'(t)^2+y'(t)^2} dt \le \int_0^{T_\infty} \sqrt{x'(t)^2+y'(t)^2} dt $$
What can I do to finish ? 
Thank you.
 A: This picture illustrates my idea from comment. Map $h$ denotes the conjugating homeomorphism between linear and non-linear flows and $G$ is some closed domain.

As you've already proved, there's a neighbourhood of saddle which contains no periodic orbits. Also, any periodic orbit that is very close to saddle has non-empty intersection with $h(G)$. Combining these two facts,we get that periodic orbit necessarily leaves $h(G)$ and its period is greater or equal to the amount of time spent in $h(G)$. But, if trajectory $\gamma$ leaves $h(G)$ then its preimage $h^{-1}(\gamma)$ (where it is defined, of course) leaves $G$. From conjugacy also follows that amount of time that $h^{-1}(\gamma)$ spends in $G$ is the same.
It's easy to derive this time for nice choice of $G$. For example, you can choose $G = \lbrace (x, y) \colon \vert x \vert < \delta, \vert y \vert < \delta \rbrace$. Suppose that periodic orbit enters through point $(\delta, y_0)$, $y_0 > 0$. By $\tau (y_0)$ denote the time at which trajectory intersects line $y = \delta$. Since the equations for linearized system are uncoupled, the solution for $y$ that passes through $(\delta, y_0)$ is $y(t) = y_0 e^{\alpha t}$ (for some $\alpha > 0$). The time $\tau(y_0)$ equals $\frac{1}{\alpha} \ln \frac{\delta}{y_0}$. Clearly, $\tau(y_0)$ tends to $+\infty$ as $y_0$ tends to zero.
