$\quad$I am recently reading Proof of the Gradient Conjecture of R. Thom by Kurdyka, Mostowski, and Parusinski and I have a question about how to apply the curve selection lemma (CSL).
Curve Selection Lemma: Let $X\subset\mathbb{R}^n$ be a semi-analytic set and suppose $0\in \overline{X\setminus\{0\}}$. Then there exists a real-analytic curve $\gamma:[0,1)\to\mathbb{R}^n$ such that $\gamma(0)=0$ and $\gamma((0,1))\subset X\setminus \{0\}$.
$\quad$Let me first introduce the setting of the paper. Let $U$ be an open set containing the origin and $f:U\subset\mathbb{R}^n\to\mathbb{R}$ be an analytic function. Moreover, $f(0)=0$ and $\nabla f(0)=0$. For any $\varepsilon>0$, define $$W^{\varepsilon}=\{x\in U\, |\, |\nabla f(x)|^2\leq (1+\varepsilon^{-2})|\partial_r f(x)|^2\ \text{and}\ f(x)\neq 0\}.$$
In Lemma 4.1 of the paper, the authors want to show that there exists a constant $c>0$ such that for all $x\in W^{\varepsilon}$, \begin{align} |f(x)|\geq c|x|^{(1-\rho)^{-1}}, \tag{1} \end{align}
where $\rho\in (0,1)$ is an exponent such that the Lojasiewicz gradient inequality holds. In the proof, the authors say that to prove (1), it suffices to show that for any analytic curve $\gamma:[0,1)\to \mathbb{R}^n$ with $\gamma(0)=0$ and $\gamma((0,1))\subset W^{\varepsilon}$, it holds $$ f(\gamma(t))/|\gamma(t)|^{(1-\rho)^{-1}}\geq c. $$
for all $t\in (0,1)$ and for some $c>0$.
$\quad$I don't understand how to obtain this reduction from the CSL. To be precise, what is the sub-analytic set to which the CSL is applied? My (unsuccessful) attempt is the following.
Assume (1) fails. There exists a sequence $x_j\in W^{\epsilon}$ such that $$ |f(x_j)|/|x_j|^{(1-\rho)^{-1}}\leq 1/j .$$ Suppose we further know that $x_j$ converges to the origin. Under these assumptions, does the CSL ensures the existence of a real analytic curve $\gamma(t)$ as above such that $$\lim_{t\to 0} f(\gamma(t))/|\gamma(t)|^{(1-\rho)^{-1}}=0? $$
Any help or insight is highly appreciated.