Show that $f$ with $f(\overline{x})=0$ is continuous for every $\overline{x}\in[0,1]$. (a) Let $f: [0,1]\rightarrow[0,\infty[$ be a mapping such that $\forall\epsilon>0$ the set $A_{\epsilon}=\{x\in[0,1]:f(x)\geq\epsilon\}$ is finite. Show that if  $f(\overline{x})=0$, then $f$ is continuous at $\overline{x}$ .

I've been given this exercise explicitly without us being unable to solve it, since the necessary tools for it weren't introduced yet  in the lecture (I'm a first year student and we're running out of time so examples and exercises are being offloaded to the practice sessions).
However I would still like to get some understanding on what is being asked here and how a possible solution to the exercise would look like. Would anybody be so kind to point me in the right directions?

In case it is relevant, there is also a second exercise (b) that probably refers back to this one:
(b) Let there be $f:[0,1]\rightarrow[0,\infty[, f(x)=\begin{cases}
\frac{1}{q} &,\text{for }x=\frac{p}{q}\in\mathbb{Q} \text{ in simplest form}\\
0 &,\text{else.}
\end{cases}$
Show for $\overline{x}\in[0,1]$ that $f$ is continuous in $\overline{x}$ if and only if $\overline{x}\in[0,1]\setminus\mathbb{Q}$.
 A: In a) you have to show: If $f(\overline{x}) = 0$ then $f$ is continuous in $\overline{x}$.
I'll give you a scetch:
To show this use the $\varepsilon$-$\delta$ - definition of continuity and fix $\varepsilon > 0$ then $A_\varepsilon$ is finite hence can be written as $A_\varepsilon = \{x_1,\ldots,x_n\}$, then choose $$\delta < \min_{i \in \{1,\ldots,n\}} \{|\overline{x} - x_i|\}$$ and you are done.
Now prove b) by showing the equivalent statements:
i) $\overline{x}\in[0,1]\setminus\mathbb{Q}$ then $f$ is continuous in $\overline{x}$. Hint: Use a) for this one
ii) $\overline{x}\not\in[0,1]\setminus\mathbb{Q}$ then $f$ is not continous in $\overline{x}$ (pretty simple by observing that in each ball around $\overline{x}$ there are elements of $[0,1]\setminus\mathbb{Q}$).
A: Your folmulation of question (a) doesn't seem right. Here is what I think you meant:
Let $f:[0,1] \to [0,\infty]$ be a mapping such that for every $\varepsilon > 0$ we have that $A_{\varepsilon} = \{x \in [0,1] : f(x) > \varepsilon\}$ is finite. then $f$ is continuous at every point where it vanishes, i.e. $f(x) = 0$ implies that $f$ is continuous at $x$.
proof:
Let $x \in [0,1]$ such that $f(x) = 0$.
Let $\varepsilon > 0$.
since $A_{\varepsilon}$ is finite, and $x \notin A_{\varepsilon}$, we can conclude that there is some neighborhood of $x$, say $D=(x-\delta, x+\delta) \subseteq [0,1]$ such that there are no points of $A_{\varepsilon}$ in $D$ (this isn't true if $x=0$ or $x=1$, but the argument is easily fixed).
Thus, there is a $\delta > 0$ such that $|x'-x| < \delta$ implies $x' \notin A_{\varepsilon}$, which implies $0 \le f(x') < \varepsilon$, therefore $|f(x')-f(x)|<\varepsilon$.
Thus, $f$ is continuous at $x$.
hints for b:  
For the first implication, that $f(x)=0$ implies that $f$ is continuous at $x$, it suffices to prove that $f$ satisfies all the conditions of (a).
For the second implication, that $f(x) \ne 0$ implies that $f$ is not continuous at $x$, notice that $f$ vanishes at every irrational number, and that irrationals (that are contained in $[0,1]$) are dense in $[0,1]$
