This question can be seen as a continuation of my question at Convergence of the probabilities that drifted Brownian motion with jump never hits zero

Let $(W_t)_{t\ge 0}$ be a standard Brownian motion and define processes

$$X^n_t:=2+t+W_t-\ell^n(t) \quad \mbox{and} \quad X_t:=2+t+W_t-\ell(t),\quad \mbox{for all } t\ge 0,$$

where $(\ell^n)_{n\ge 1}$ and $\ell$ are right-continuous and non-decreasing functions s.t. $\ell^n(0)=\ell(0)=0$ and $0\le \ell^n(t), \ell(t)\le 1$ for all $t\ge 0$. If $\lim_{n\to\infty}\ell^n(t)=\ell^n(t)$ holds for all the points of continuity of $\ell$, can we prove

$$\lim_{n\to\infty}\mathbb P[\tau^n=\infty]=\mathbb P[\tau=\infty]?$$

Here $\tau^n:=\inf\{t\ge 0:~ X^n_t\le 0\}$ and $\tau:=\inf\{t\ge 0:~ X_t\le 0\}$.

**Personal thoughts :** My idea is the following. Take a sequence $(T_m)_{m\ge 1}$ diverging to $\infty$ s.t. $\ell$ is continuous at every $T_m.$ Then

$$\big|\mathbb P[\tau^n=\infty]-\mathbb P[\tau=\infty]\big|\le \big|\mathbb P[\tau^n>T_m]-\mathbb P[\tau^n=\infty]\big|+\big|\mathbb P[\tau^n>T_m]-\mathbb P[\tau>T_m]\big|+\big|\mathbb P[\tau>T_m]-\mathbb P[\tau=\infty]\big|.$$

If we are able to show the first and third terms can be uniformly small as $m\to\infty$, i.e. for any $\epsilon>0$, there exists $m_{\epsilon}$ s.t.

$$ \big|\mathbb P[\tau^n>T_m]-\mathbb P[\tau^n=\infty]\big|+\big|\mathbb P[\tau>T_m]-\mathbb P[\tau=\infty]\big|\le \epsilon,\quad \mbox{for all } m\ge m_{\epsilon}.\quad \quad (\ast)$$

Then it suffices to show for fixed $m_{\epsilon}$, one has

$$\lim_{n\to\infty}\big|\mathbb P[\tau^n>T_{m_{\epsilon}}]-\mathbb P[\tau>T_{m_{\epsilon}}]\big|=0.$$

But I don't know how to prove $(\ast)$.