A Question on Borel Measurable Functions

I'll do (i) for $A$ and leave to you the specular argument for $B$.

The proof I propose can be replicated in any metric space, therefore I'll mix the general notation with the one of the real line.

Call $\mathcal B(x,r)$ the open ball centered in $x$ with radius $r$. As already pointed out, if we define, for $\varepsilon>0$ and $x\in[a,b]$, $$g_\varepsilon(x)=\sup\{f(y)\,:\,y\in \mathcal B(x,\varepsilon)\cap[a,b]\}$$ (these real numbers are well defined because $f$ is bounded) then the family of functions $(g_\varepsilon\,:\,\varepsilon>0)$ is an increasing family, id est, $$\varepsilon<\delta\implies\forall x,\ g_\varepsilon(x)\le g_\delta(x)$$ because of the fact that $\mathcal B(x,\delta)\cap[a,b]\supseteq\mathcal B(x,\varepsilon)\cap[a,b]$.

So, in order to prove that the decreasing limit $A(\bullet)=\lim_{\varepsilon\to 0^+} g_\varepsilon(\bullet)$ is a Borel function, you only need to prove that each $g_\varepsilon$ is Borel. But each of these functions is actually lower semi-continuous on the interval $[a,b]$, id est, for all $\alpha$ the set $g^{-1}_\varepsilon(\alpha,\infty):=\{x\in [a,b]\,:\, g_\varepsilon(x)>\alpha\}$ is an open subset of $[a,b]$.

In fact let $x\in g_\varepsilon^{-1}(\alpha,\infty)$. $$\sup\{f(y)\,:\,y\in \mathcal B(x,\varepsilon)\cap[a,b]\}>\alpha\iff \exists y'\in\mathcal B(x,\varepsilon)\cap[a,b],\ f(y')>\alpha$$

Since $\lvert x-y'\rvert<\varepsilon$, I claim that $\mathcal B(x,\varepsilon-\lvert x-y'\rvert)\cap[a,b]\subseteq g_\varepsilon^{-1}(\alpha,\infty)$. In fact, let $z\in\mathcal B(x,\varepsilon-\lvert x-y'\rvert)\cap[a,b]$. Then, $y'\in\mathcal B(z,\varepsilon)$, since $$\lvert z-y'\rvert\le\lvert z-x\rvert+\lvert x-y'\rvert<\varepsilon-\lvert x-y'\rvert+\lvert x-y'\rvert$$ and by definition $f(y')>\alpha$. So, indeed $g_\varepsilon(z)>\alpha$, which proves that $g^{-1}_\varepsilon(\alpha,\infty)$ is a neighbourhood in $[a,b]$ of all its points.

For $B$, there is $\inf$ instead of $\sup$, the family $(\widehat g_\varepsilon\,:\, \varepsilon>0)$ is decreasing (hence the limit as $\varepsilon\to 0^+$ is increasing) and there is to prove that the new $\widehat g_\varepsilon$ are upper semi-continuous, id est that $\widehat g_\varepsilon^{-1}(-\infty,\alpha)$ is open.

(ii) is rather obvious, and I have not much to add to the other answer.

Now, to (iii). As you've already proven, $B(x)\le f(x)\le A(x)$ always holds. Suppose $A(x_0)=f(x_0)=B(x_0)$ and assume as a contradiction that $f$ is not continuous at $x_0$. Since $[a,b]$ is a metric space, there must be a sequence $y_n\to x_0$ and a fixed $\delta>0$ such that for all $n\in\Bbb N$, $f(y_n)\in (-\infty,f(x_0)-\varepsilon)\cup(f(x_0)+\varepsilon,+\infty)$. Up to extracting a subsequence, we can assume that $f(y_n)$ lies in either of the two components. If, say, $f(y_n)\in(f(x_0)+\varepsilon,+\infty)$ for all $n$, this means that $\sup\{f(y)\,:\, \lvert x_0-y\rvert<\varepsilon\wedge y\in[a,b]\}\ge f(x_0)+\varepsilon$ for all $\varepsilon>0$, since we have the points $y_n\to x_0$. Hence, $A(x_0)\ge f(x_0)+\varepsilon$, which is absurd. Again, a specular argument works with $B$ in case $f(y_n)\in(-\infty,f(x_0)-\varepsilon)$.