$\def\dts{\mathinner{\ldotp\ldotp}}$ First, we extend the domain $[0\dts1]$ into $\Bbb R$, say, $f(x_0)=\lim_{x\to0^+}f(x)$ whenever $x_0<0$, and $f(x_0)=\lim_{x\to1^-}f(x)$ whenever $x_0>1$, hence we can conclude that $\lim_{x\to a}f(x)$ exists for each real number $a$. Just like the notation in (ii), and for ease, let $g(x)=\lim_{y\to x}f(x)$, so $f^c(x)=g(x)$ on the interval $[0\dts1]$.
Next, we can say that $f$ is bounded.
By definition of limit of function, there exists $\delta_\epsilon(a)>0$ such that $|f(x)-g(a)|<\epsilon$ whenever $0<|x-a|<\delta_\epsilon(a)$, therefore $|f(x)|\le\max(|f(a)|,|g(a)|+\epsilon)=M_\epsilon(a)$ whenever $|x-a|<\delta_\epsilon(a)$.
Supposing that $\epsilon$ is a fixed positive real number, say $1$, and let $\Sigma$ is the set of open intervals $I(x)=(x-\delta_\epsilon(x)\dts x+\delta_\epsilon(x))$, where $0\le x\le1$. We have $\Sigma_\epsilon$ covers the closed set $[0\dts1]$, thus there exists finite subset of $\Sigma_\epsilon$, say $\Sigma_\epsilon^*$, which covers the closed set $[0\dts1]$, and let $\Sigma_\epsilon^*=\{I_\epsilon(a_1),\ldots,I_\epsilon(a_m)\}$, we have $|f(x)|\le M_\epsilon(a_k)$ on the open interval $I_\epsilon(a_k)$, so $f$ is bounded.
Now, let's prove that $g$ is continuous. For all $\epsilon>0$, we have $|f(x)-g(x_0)|<\epsilon$ where $x$ is an arbitrary point on the open interval $I_\epsilon(x_0)$. Let $y\in I_\epsilon(x_0)$, and let $x$ tends to $y$ but not equals $y$, we have $f(x)\to g(y)$, therefore $|g(y)-g(x_0)|\le\epsilon$, hence $g$ is continuous.
Then, supposing that $f^c(x)=0$ on the interval $[0\dts1]$, we have $g(x)=0$ for arbitrary real number $x$. For all $\epsilon>0$, we should remember that there's a corresponding finite cover $\Sigma_\epsilon^*$ of closed set $[0\dts1]$, and $\Sigma_\epsilon^*=\{I_\epsilon(a_1),\ldots,I_\epsilon(a_m)\}$. Now let's analyse this result closely. For $x\in I_\epsilon(a_k)$, and $x\neq a_k$, we have $|f(x)|=|f(x)-g(a_k)|<\epsilon$, so $|f(x)|\ge\epsilon\,\Longrightarrow\,x=a_k\textrm{ for some }k$. Thus we can sort all $x$ such that $f(x)\neq0$ in decreasing order of $|f(x)|$, and arrange them in a sequence.
Finally, we should prove that, if we can arrange all non-zero points on interval $[0\dts1]$ in a sequence $S$ such that the value converges to $0$, we have $f^c=0$. For any point $y$ in $[0\dts1]$ and arbitrary positive real number $\epsilon$, there's only finity many points in $S$ whose absolute value $\ge\epsilon$, thus $f(x)<\epsilon$ holds on some neighborhood of $y$ (if $y=0,1$, it's one sided neighborhood), so $f^c(y)=0$.