Use the definition $\epsilon - \delta$ of continuity for proof that if the function $f: \mathbb{R}\longrightarrow \mathbb{R}$ is continuous in a, then $f$ is bounded in an open interval centered on a.
My idea for this exercise is the next. Let $\epsilon >0$. We now that $f$ is a continuous function in a, i.e., exists $\delta>0$ such that $$ |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon. $$ Then for properties of abs value, $$ |x-a|<\delta \Rightarrow -\epsilon+f(a)<f(x)<\epsilon +f(a) $$ $$ |x-a|<\delta \Rightarrow f(x)\in ]-\epsilon+f(a),\epsilon +f(a)[ $$ and well, $]-\epsilon+f(a),\epsilon +f(a)[$ is an open interval centered on $f(a)$, but the exercise said me an open interval centered on a, and I don't know what I have to do for the next steps.
You are (perhaps reasonably) misinterpreting the intended significance of the phrase :
$f$ is bounded in an open interval centered on $a$.
What the statement intends is that there exists an $M > 0$ and a $\delta > 0$ such that
for all $x$, where $a - \delta < x < a + \delta$, you will have that $|f(x)| < M.$
The statement that you are misinterpreting is badly worded, but that is the intent.
The easy approach is to take $\epsilon = 1.$
Then, there exists some $\delta > 0$ such that for all $x$ where $a - \delta < x < a + \delta$, by the definition of continuity, you will have that
$$f(a) - 1 < f(x) < f(a) + 1.$$
This justifies the assertion that there exists a $\delta > 0$, such that $f(x)$ is bounded, for all $x$ such that $a - \delta < x < a + \delta$.