If $f(a^+)$ and $f(a^-)$ exist, then $f$ is bounded.

Such an $f$ is called a regulated function. The space of step functions is dense in the space of regulated functions for $\|\cdot\|_{\infty}$, so $\exists \phi$ a step function $[0,1] \to \Bbb R$ such that $\|\phi - f\|_{\infty} \le 1$. Thus $\|f\|_{\infty} \le 1 + \|\phi\|_{\infty} < \infty$.

For $B$, consider the function:

$$f(x) = \begin{cases} x, 0\le x < 1/2 \\ 0, 1/2 \le x \le 1 \end{cases}$$

For $C$, WLOG say $f(0) < 0$ and $f(1) > 0$. As $f$ is continuous at $0$, $\exists 0<\delta < 1$ such that $f(x) < 0$ for all $0\le x \le \delta$. Similarly, there is $\epsilon \in (0,1)$ such that $f(x) > 0$ for all $x \in [1-\epsilon, 1]$.

Let: $$S =\{s \in [0,1]: \text{for all but finitely many } t\in [0,s], f(t) < 0\}$$

Let $p = \sup S$. We have $0<p<1$ since $\delta \in S$ and $S \subset [0,1-\epsilon]$. Clearly $f(p-) < 0$. Now suppose that $f(p+)<0$. We have $f(x) < \frac{f(p+)}{2} < 0$ for all $x \in (p, p+\eta)$ for some small $\eta > 0$. It follows, thus, that $p+\eta/2 \in S$, so $p+\eta/2 \le p$, a contradiction. Therefore $f(p+) \ge 0$, and we get $f(p-)f(p+) \le 0$.

A) Let $x_n$ be such that $|f(x_n)|\to \sup |f|$. By compactness, there exists a convergent subsequence, and furthermore, it can be chosen to be monotone. By assumption, the limit along the monotone subsequence exists. Thus $\sup |f|$ is finite.

B) Answered before.

C) WLOG $f(0)<0$. Then by continuity at $0$ and $1$, there exists $\epsilon\le \frac 12$ such that $f(p-)<0$ for $p$ in $[0,\epsilon)$ and $f(p-)>0$ for $p\in (1-\epsilon,1]$. Let $\bar p =\sup\{p:f(p-)<0\}$. Then ${\bar p} \in [\epsilon,1-\epsilon]$. By assumption, for every $n$, there exists $p_n- \in (\bar p-\frac 1n,\bar p)$ such that $f(p_n-)< 0$ and $p_n+\in (\bar p , \bar p + \frac 1n)$ such that $f(p_n)\ge 0$. Therefore $f(p-)= \lim f(p_n-)\le 0$, while $f(p+)=\lim f(p_n+)\ge 0$.