Could you please verify my attempt is correct or contains mistakes? Thank you so much!
Let $I=[a,b] \subseteq \mathbb R$ and $F$ a Banach space. Then $f:I \to F$ is jump continuous if and only if there is a sequence of staircase functions converging uniformly to $f$.
My attempt:
$\Rightarrow$ Suppose $f:I \to F$ is a jump continuous function. Then for all $x \in I$, there are numbers $L(x)$ and $R(x)$ such that $$\| f(s)- f(t) \| < 1/n \quad \text{for} \quad s,t \in (L(x),x) \cap I \quad \text{or} \quad s,t \in (x,R(x)) \cap I$$
Because $\{(L(x),R(x)) \mid x \in I\}$ is an open cover of the compact interval $I$, there are $x_1, \ldots, x_n$ such that $\bigcup_{i=1}^n (L(x_i),R(x_i))$ covers $I$. Let $\mathfrak{Z} = (\xi_1,\ldots,\xi_m)$ be a partition of $I$ generated by $$\{a,b\} \cup \bigcup \{L(x_i),x_i, R(x_i) \mid i=\overline{1,n}\}$$
Define $f_n: I \to F$ by $$f_n(x) = \begin{cases} f(x) & \text{if} \quad x \in \{\xi_i \mid i = \overline{1,m}\} \\ f((\xi_i + \xi_{i+1})/2) & \text{if} \quad x \in (\xi_{i-1} , \xi_{i}) \, \, , \quad i = \overline{2,m} \end{cases}$$ Then $f_n$ is a staircase function, and by construction $$\forall x \in I:\|f_n(x) - f(x)\| < 1/n$$ Therefore, $\| f_n -f \|_\infty \le 1/n$.
$\Leftarrow$ Suppose the sequence $(f_n)$ of staircase functions converges uniformly to $f$. For $\varepsilon>0$, there is $n \in \mathbb N$ such that $\|f_n (x) - f(x)\| < \varepsilon/2$ for all $x \in I$. Moreover, for every $x \in (a,b]$, there is $\alpha \in [a,x)$ such that $f_n$ is constant on $(\alpha,x)$. Consequently, $$\|f(s)-f(t)\| \leq\left\|f(s)-f_{n}(s)\right\|+\left\|f_{n}(s)-f_{n}(t)\right\|+\left\|f_{n}(t)-f(t)\right\| < \varepsilon$$ for all $s,t \in (\alpha,x)$.
Suppose now $(s_{i})$ is a sequence in $I$ that converges from the left to $x$. It follows from the above inequality that $(f(s_i))$ is a Cauchy sequence in the Banach space $F$. Thus $f(s_i) \to y$ as $s_i \to x$. Suppose there is another sequence $(t_{j})$ in $I$ that converges from the left to $x$. By the same argument, we get $f(t_j) \to y'$ as $t_j \to x$.
There is $N \in \mathbb N$ such that $s_i,t_j \in (\alpha,x)$ for all $i,j \ge N$. Hence $\|f(s_i)-f(t_j)\|< \varepsilon$ for all $i,j \ge N$. Take the limit $i \to \infty$, we get $\|y-f(t_j)\|< \varepsilon$ for all $j \ge N$. Because we can make $\varepsilon$ arbitrarily small, $f(t_j) \to y$. Hence $y=y'$ and thus $\lim_{s \to x^-} f(s)$ exists.
By swapping left and right, we show that for $x \in[a, b)$ the limit $\lim_{s \to x^+} f(s)$ exists as well. Consequently, $f$ is jump continuous.