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If $f$ is continuous at $x_0$, choose a small interval on which $|f(s)-f(x_0)|<ϵ/2$. The triangle inequality makes its oscillation less than $ϵ$, so $\phi (x_0)=0$.

Conversely, if $\phi (x_0)=0$, then for every $ϵ>0$ some $n$ has ${\phi }_n(x_0)<ϵ$. As $x_0$ itself belongs to the interval used in that supremum, $|f(x)-f(x_0)|<ϵ$ for $|x-x_0|<1/n$. Thus $f$ is continuous at $x_0$.

The continuity set of a function is thereforeEach set in the intersection is open by part (i), so this is a G-delta set and in particular a Borel set, even if $f$ itself is not measurable.

Yes. In fact, measurability of $f$ is unnecessary. Define the oscillationFor every $\epsilon >0$, the set $\{x:{\omega }_f(x)<\epsilon \}$ is open, and $f$ is continuous at $x$ exactly when ${\omega }_f(x)=0$. Hence the continuity set of a function isa $G_{\delta }$ set and therefore a Borel set.