Why the stable manifold theorem and the Hartman-Grobman theorem implies saddles to be unstable? Perko's book states that any sink $x_0\in E\subset\mathbb{R}^n$ of 
\begin{equation}\label{uno}
\dot{x}=f(x),
\end{equation}
where $x\in E\subset\mathbb{R}^n$ and $f\in C^1(E)$, is asymptotically stable and that any saddle or source of this differential equation is unstable, as a consequence of the Stable Manifold Theorem and the Hartman-Grobman Theorem.
I understand that a sink must be asymptotically stable by the Hartman-Grobman Theorem, since $\dot{x}=f(x)$ has the same qualitative structure than $\dot{x}=Df(x_0)(x-x_0),$ being $x_0$ the equilibria point. The same argument follows for a source. However, it is not clear to me why this also follows for a saddle.
My idea for proving the instability of a saddle is that as there is a part of the unstable manifold in every vicinity of $x_0$, any point $x$ on it is unbounded in the sense that its flow $\phi_t(x)$ goes farther as one wants from $x_0$, for an appropriate $t\geq0$. Of course, I haven't been able to prove this.
I look for an explanation that only involves the Stable Manifold Theorem and the Hartman-Grobman Theorem, like the stable curve theorem in Hirsch and Smale's book.
 A: After thinking about it, I noticed that if $x_0$ is a saddle of
$$\dot{x}=f(x),$$
by the stable manifold theorem, there is a $k$-manifold $U$, where $k$ is the number of eigenvalues with negative part, such that any point $x\in U$ satisfies that
$$\lim_{t\rightarrow-\infty}\phi_t(x)=x_0.$$
Now, to show that $x_0$ is not stable, choose $\epsilon>0$ such that the open ball $B_\epsilon(x_0)$ (centred in $x_0$, with radius $\epsilon$) is fully contained in the open set where the stable manifold theorem is valid, and also such that $B_\epsilon(x_0)^c\cap U\neq\emptyset$, in order to take a point $y\in U$ such that $y\not\in B_\epsilon(x_0)$. 
Since $y\in U$, $\forall\delta>0, \ \exists N>0\text{ such that } |\phi_{-n}(y)-x_0|<\delta,$ for $n>N.$ This implies that the saddle point is not stable, because there exists $\epsilon>0$ such that for all $\delta>0$, exists $\phi_{-n}(y)\in B_\delta(x_0)$, with $n>N$, such that
$$\phi_n(\phi_{-n}(y))=y\not\in B_\epsilon(x_0).$$
To resume, in any vicinity of a saddle $x_0$, one can always find a point (in the unstable manifold) such that its flow goes farther as one wants from $x_0$ (inside the region where the theorem is valid).
