Let $f,f_n:[0,\infty)\to\mathbb R$ be continuous for $n\in\mathbb N$ with $$\sup_{s\in[0,\:t]}|f_n(s)-f(s)|\xrightarrow{n\to\infty}0\;\;\;\text{for all }t\ge0,\tag1$$ $C\ge0$, $$\tau_n:=\inf\left\{t\ge0:|f_n(t)|\ge C\right\}\;\;\;\text{for }n\in\mathbb N$$ and $$\tau:=\inf\left\{t\ge0:|f(t)|\ge C\right\}$$ with the convention $\inf\emptyset=\infty$.
How can we show that $$f_n(\tau_n\wedge t)\xrightarrow{n\to\infty}f(\tau\wedge t)\tag2\;\;\;\text{for all }t\ge0,$$ where $x\wedge y:=\min(x,y)$?
Let $$g_n(t):=\sup_{s\in[0,\:t]}|f_n(s)|\;\;\;\text{for }t\ge0$$ for $n\in\mathbb N$ and $$g(t):=\sup_{s\in[0,\:t]}|f(s)|\;\;\;\text{for }t\ge0.$$ Let's consider the case $0<\tau<\infty$. We can show that $$|f(\tau)|=C\tag3.$$ By the reverse triangle inequality and $(1)$, $$|g_n(\tau)-g(\tau)|\le\sup_{s\in[0,\:\tau]}|f_n(s)-f(s)|\xrightarrow{n\to\infty}0\tag4.$$ Let $\varepsilon>0$. By $(3)$ and $(4)$, there is a a $N\in\mathbb N$ with $$|g_n(\tau)-C|<\varepsilon\;\;\;\text{for all }n\ge N\tag5.$$ My hope is that we can somehow conclude $\tau_n\xrightarrow{n\to\infty}\tau$ from $(5)$, but I wasn't able to figure out how.