If $\sup_{t∈[0,\:t_0]}\left\|x^n_t-x_t\right\|_E\xrightarrow{n→∞}0$ for all $t_0≥0$, then $\sup_{t≥0}\left\|x^n_t-x_t\right\|_E\xrightarrow{n→∞}0$

Question

Let $E$ be a normed space, $(x_t)_{t\ge 0}\subseteq E$ with $$\sup_{t\in[0,\:t_0]}\left\|x_t\right\|_E<\infty\;\;\;\text{for all }t_0\ge 0\tag 1$$ and $(x^n_t)_{t\ge 0}\subseteq E$ with $$\sup_{t\ge 0}\left\|x^n_t\right\|_E<\infty\;\;\;\text{for all }t_0\ge 0\tag 2$$ for $n\in\mathbb N$. Assume that $$\sup_{t\in[0,\:t_0]}\left\|x^n_t-x_t\right\|_E\xrightarrow{n\to\infty}0\;\;\;\text{for all }t_0\ge 0\;.\tag 3$$ I want to show that $$\sup_{t\ge 0}\left\|x^n_t-x_t\right\|_E\xrightarrow{n\to\infty}0\;.\tag 4$$ Is that possbile? And if so, how?