Topology containing a parabolic region 
We consider in $\mathbb{R}^2$ the topology of the sets $U\subset \mathbb{R}^2$ which contains for each $(a,b)\in U$ a parabolic region given by 
  $$\alpha(x-a)^2 <y-b, \quad \alpha >0$$
  Study if this space verifies the first-countable axiom, the second-countably axiom and if it's separable. Also, show that there are compact set in this space which are neither closed nor bounded. 

Firstly, we will think about how the open sets are. For each point of the plane, we can take $\alpha>0$ and the parabolic region 
$$\alpha(x-a)^2 <y-b,$$
and observe that $(a,b)$ does not satisfy the equality. Changing $\alpha$ will widen or narrow the parabolic region. Then, an open set in this topology will be 
$$U=\{(a,b)\}\cup \{(x,y)\in \mathbb{R}^2 : \alpha(x-a)^2 <y-b\}.$$
If we fix $\alpha$ for each point $(x,y)$ of the parabolic region (remember that the set $U$ has to contain $\textbf{a}$ parabolic region), we will be able to choose $\alpha_{(a,b)}$ (it can be very big) such that $U$ will contain all of the parabolic region with vertex $(x,y)$. 
Now, we can work on the problem. 
$(a)$ The set $\mathbb{Q}^2$ intersects each open set (non-empty), so the space is separable. 
$(b)$ Taking $\beta>\alpha_{(a,b)}$, where $\beta\in \mathbb{N}$, we have that
$$\beta(x-a)^2 <\alpha_{(a,b)}(x-a)^2 <y-b.$$
Thus, we can take as a neighbourhood basis for the point $(a,b)$
$$\mathcal{V}_{(a,b)}=\{U_{a,b}^\beta: \beta\in \mathbb{N}, \beta > \alpha \}$$
where 
$$U_{a,b}^\beta=\{(a,b)\}\cup \{(x,y)\in \mathbb{R}^2 : \beta(x-a)^2 <y-b\}.$$
This basis is countable, so the space is first-countable axiom. 
$(c)$ Given an open set $U$, then we intersect it with the line $r: y=b$. The intersection is just the point $(a,b)$ and, by the properties of relative topology in the line $r$, we have that the point $(a,b)$ is open in $(r, \mathcal{T}|_r)$. As $\mathbb{R}^2$ is not countable, then we cannot find a countable basis of open sets, so this space is not second-countably axiom. 
$(d)$ We consider the set 
$$K=\{a\}\times [b+1,+\infty)$$
and an open covering of $K$. It's clear that there is $\alpha>0$ such that 
$$K\subset \{(a,b)\}\cup \{(x,y)\in \mathbb{R}^2 : \alpha(x-a)^2 <y-b\},$$
which is finite. Then, $K$ is a compact neither closed nor bounded. 
 A: Let $P(a,b,\alpha) = \{(x,y) \in \mathbb{R}^2 \mid \alpha(x-a)^2 <y-b \} \cup \{ (a,b) \} $
Possible interpretations of the definition of openness are:
$U$ is open iff $\forall (a,b) \in U, \exists \alpha > 0, P(a,b,\alpha) \subset U$
$U$ is open iff $\forall (a,b) \in U, \forall \alpha > 0, P(a,b,\alpha) \subset U$
$U$ is open iff $\exists \alpha > 0, \forall (a,b) \in U,  P(a,b,\alpha) \subset U$
The first is the most natural, but the other two do determine topologies. Actually they define the same topology. For each point (a,b) let $H(a,b) = \{ (x,y) \in \mathbb{R}^2 \mid y > b \} \cup \{(a,b) \} $. This is a nbhd of (a,b) in the latter topology and is contained in any other nbhd of (a,b). Hence the topology is first countable but not second countable. It is clearly separable. Any non-empty compact subset has to be bounded below and have finitely many points (but not none) at the lowest level and any such set is compact. None are closed.
If the first definition is used, then each $P(a,b,\alpha)$ is open - for any other point (c,d) in the set, a large enough $\beta$ gives $P(c,d,\beta) \subset P(a,b,\alpha)$. Restricting $\alpha$ to positive naturals gives a countable nbhd basis. But as before the topology is separable but not second countable. Any non-empty compact subset is covered by finitely many of these Ps and so is bounded below. $\{ (x,y) \in \mathbb{R}^2 \mid y \geq b \}$ is compact but not bounded above.
ETA. Your rewrite of the question seems to have reached the same answer.
A: You have reached the right answers, but your explanations are rather confused. I've added comments after passages copied from your post. (I'm afraid the quotes don't reproduce very well. I don't know how to get round that.)
"Then, an open set in this topology will be
U={(a,b)}∪{(x,y)∈R2:α(x−a)^2 < y-b}"
No, an open set U contains a set of this form for each of its points (a,b). It needn't be of this form itself, and you haven't yet proved such sets are themselves open.
"If we fix α for each point (x,y) of the parabolic region (remember that the set U has to contain a parabolic region), we will be able to choose α(a,b) (it can be very big) such that U will contain all of the parabolic region with vertex (x,y)"
This is almost how to prove such sets are open, but reverses the logic. You need to start with a particular (a,b) and $\alpha$ and then show that for each other (x,y) in the set you can choose an $\beta$ so that the parabolic region above (x,y) with that parameter is contained within the original one above (a,b). (To prove that the sets defined as open do actually form a topology we also need to show that the intersection of any finite collection of them is open. That is straightforward but, since the exam question tells you it's a topology I suppose they don't require you to prove it.)
...
"(b)
Taking β>α(a,b), where β∈N, we have that
β(x−a)2<α(a,b)(x−a)2α}
where
Uβa,b={(a,b)}∪{(x,y)∈R2:β(x−a)2 "
There is no single $\alpha_{a,b}$. Different values of $\alpha$ will be needed for different open sets containing (a,b). Once you've shown that each $U_{a,b}^\beta$ is open it follows immediately that they form a  nbhd basis for (a,b). Leave out the $\beta > \alpha$ in the definition of $\mathcal{V}_{(a,b)}$.
"(c)
Given an open set U, then we intersect it with the line r:y=b. The intersection is just the point (a,b) and, by the properties of relative topology in the line r, we have that the point (a,b) is open in (r,T|r)."
An arbitrary open set can intersect the line r in any possible subset. The point is that for any (a,b) on that line there is an open set ($U_{a,b}^\beta$ for any $\beta$) which contains (a,b) but no other point on the line.
"(d)
We consider the set
K={a}×[b+1,+∞)
and an open covering of K. It's clear that there is α>0 such that
K⊂{(a,b)}∪{(x,y)∈R2:α(x−a) 
, which is finite."
This doesn't make much sense. $\{ (a,b) \} \cup \{(x.y) \in \mathbb{R}^2 \mid \alpha(x-a)^2 < y-b \}$ is not finite. The collection containing just that set is finite and does cover K but it is not a subcovering of the given covering. Just note that the covering must include an open set which contains (a,b+1) and which - by the original definition - must therefore be a superset of K. So we have a subcovering with one member.
