The (a) part is clear:
Suppose that $f$ is not continuous at $a$. Prove that for some number $\varepsilon>0$ there are numbers $x$ arbitrarily close to $a$ with $|f(x)-f(a)|>\varepsilon$.
It's not the case that $f$ continuous. Therefore it's not the case that $$\forall\varepsilon>0,\exists\delta>0,\forall x(|x-a|<\delta \rightarrow |f(x)-f(a)|<\varepsilon)$$
which is logically equivalent to
$$\exists\varepsilon>0,\forall\delta>0,\exists x(|x-a|<\delta \mbox{ and } |f(x)-f(a)|\geq\varepsilon)$$
There are $\varepsilon>0$ such that $|f(x)-f(a)|>\varepsilon$. Let $\varepsilon'=\frac12\varepsilon$. Then we have $|f(x)-f(a)|\geq\varepsilon>\varepsilon'$
The (b) part is not so much:
Conclude that for some number $\varepsilon>0$ either there are numbers $x$ arbitrarily close to $a$ with $f(x)<f(a)-\varepsilon$ or there are numbers $x$ arbitrarily close to $a$ with $f(x)>f(a)+\varepsilon$
From (a) we have $|f(x)-f(a)|>\varepsilon$, which implies $f(a)-\varepsilon>f(x)>f(a)+\varepsilon$.
But $(f(a)+\varepsilon, f(a)-\varepsilon)=(-\infty,f(a)-\varepsilon) \cap (f(a)+\varepsilon,+\infty)$ is an empty set, $f(x)$ can't be in there.
How does it follow that that $f(x)\in (-\infty,f(a)-\varepsilon) \cup (f(a)+\varepsilon,+\infty)$?
Also it seems like a contradiction, does it mean that $f$ is actually continuous at $a$ (which is nonsense, all functions are continuous for all $a$)?
There exists a sequence of distinct points $x_n \to a$ such that $|f(x_n)-f(a)| >\epsilon$ for all $n$. The sets $\{n: f_n(x) >f(a)+\epsilon\}$ and $\{n: f_n(x) <f(a)-\epsilon\}$ cannot both be finite. (If they are both finite then $\mathbb N$ would be a finite set). If there are infinitely many points in the first set then there are points arbitrarily close to $a$ with $f(x) >f(a)+\epsilon$. If there are infinitely many points in the second set then there are points arbitrarily close to $a$ with $f(x) <f(a)-\epsilon$.