+\infty}\|u(t, \cdot)\|_{C([0, h(t)])}}= \varepsilon >0.$$ Then, there exists a sequence $(t_{k}, x_{k})$ in $(0, \infty)\times [0, h(t)]$, where $t_{k} \rightarrow \infty$ as $k \rightarrow \infty$, such that $$u(t_{k}, x_{k})\geq \frac{\varepsilon}{2}\text{ for all }k \in \mathbb{N}.$$ Note that $0\leq x_{k}<h(t_{k})< h_{\infty} < \infty.$ By passing to a subsequence if necessary, it follows that $x_{k} \rightarrow x_{0} \in (0, h_{\infty})$ as $k \rightarrow \infty$. Define $$u_{k}(t, x):=u (t+t_{k},x)~\text{ and }~v_{k}(t, x)=v (t+t_{k},x)$$ for $t \in (-t_{k}, \infty)$ and $x \in [0, h(t+t_{k})]$. It follows from and standard parabolic regularity that $\{(u_{k},v_{k}) \}$ has a subsequence $\{(u_{k_{i}},v_{k_{i}}) \}$ satisfying $(u_{k_{i}},v_{k_{i}}) \rightarrow (\tilde{u}, \tilde{v})$ as $k_{i} \rightarrow \infty$, where $(\tilde{u}, \tilde{v})$ is the solution to the following system $$\label{eqth4.2y1}
\left\{\begin{array}{ll}
\displaystyle{\frac{\partial \tilde{u}}{\partial t}= d_{1}\tilde{u}_{xx}+\tilde{u}\left[b_{1}-\delta_{1}(\tilde{u}+\tilde{v})\right],}&(t,x)\in (-\infty, +\infty)\times(0, h_{\infty}),\vspace{7 pt}\\

\displaystyle{\frac{\partial\tilde{v}}{\partial t}=d_{2}\tilde{v}_{xx}+\tilde{v}\left[\frac{b_{2}\tilde{v}}{\tilde{u}+\tilde{v}}-\delta_{2}(\tilde{u}+\tilde{v})\right],}&(t,x)\in (-\infty, +\infty)\times(0, h_{\infty}),
\end{array}\right.$$ with $\tilde{u}(t,h_{\infty})=0$ for all $t\in\mathbb{R}$. Since $$\tilde{u}(0,x_{0})=\lim_{k_{i} \rightarrow  \infty} u_{k_{i}}(0,x_{k_{i}})=\lim_{k_{i} \rightarrow  \infty}u(t_{k_{i}},x_{t_{k_{i}}})\geq \frac{\varepsilon}{2},$$ the maximum principle implies that $\tilde{u} > 0$ in $(-\infty, +\infty)\times(0, h_{\infty})$. Hence, we can apply Hopf Lemma at the point $(0, h_{\infty})$ to obtain $\tilde{u}_{x}(0, h_{\infty})< 0.$ Therefore, we have $u_{x}(t_{k_{i}}, h(t_{k_{i}}))= \partial_{x} u_{k_{i}}(0, h(t_{k_{i}}))< 0$ for large $i$. This, together with the Stefan condition, implies that $h'(t_{k_{i}})>0$ .

On the other hand, $h_{\infty}< +\infty$ implies 