Interior and closure of subsets 
Let $X= \mathbb{R^2}$.
$\tau$ is the topology on $X$ which has the basis $\mathbb{B}= \{ (x-\epsilon,x]\times (y-\epsilon,y] : x,y\in \mathbb R\land \epsilon 
 \in \mathbb R^+\}.$
Decide if each of the following subsets are open,closed in $\tau$ and
  find the interior and the closure.
$A=[0,1]\times\mathbb{R}$
$B=$$\mathbb{R}\times(-1,1)$
$C=\{(x,y)\in  \mathbb R^2:x^2+y^2<1\}$.

I think $A$ is not an open set because I can't obtain $[0,1]$ as union of the basis elements. So $Int(A)=(0,1]\times \mathbb R$.
A is not even closed because its complement is not an open set. $Cl(A)=\mathbb  R^2$.
In $B$ I found that $Cl(B)=\mathbb R\times (-1,1]$. I think $B$ is open, therefore $Int(B)=B$, because $(-1,1)=\cup_{n=1}^{\infty}(y-\epsilon,y - 1/n)=(y-\epsilon,y)$ and for $y=1$ and $\epsilon=2,$ I get $ (-1,1)$.
I hope to be right. But I don't know how to move for $C$. 
edit
Certainly C is not open because it's a circle and I can't obtain it as union of squares. I want to write te biggest open set I can found in it. It's the square with l=2/sqrt(2).And Cl(C)=(−1,1]x(−1,1]
 A: You can't say

because I can't obtain $[0,1]$ as union of the basis element.

because that is too sloppy and you can conclude that it is impossible. The right argument is to find a boundary point $a\in A$ and show that no element of the base $\mathbb B$ which contains $a$ is contained in $A$.
Proof that $A$ isn't open: 
Consider $a=(0,0)$. Let be $U:=(x_0-\epsilon,x_0]\times (y_0-\epsilon,y_0]\in \mathbb B$ such that $a\in U$. You can conclude $y_0-\epsilon<a_2=0$ and there exists $y\in(y_0-\epsilon,0)$ such that $(0,y)\in U$. But $(0,y)\notin A$ since $y<0$ and therefore $U\not\subset A$ and $A$ isn't open because $(0,1)\in A$ but $(0,1)\notin int(A)$.$\Box$
Proof that $A$ is closed:
Consider $(x,y)\in X\setminus A$ then $y>1$ or $y<0$. 
If $y>1$ then define $U=(x-1,x]\times(1,y]\in\mathbb B$. We get $U\subset X\setminus A$ and $(x,y)\in U$.
If $y<0$ define $U=(x-1,x]\times(2y,y/2]\in\mathbb B$ and you get $U\subset X\setminus A$ and $(x,y)\in U$.
Therefore $X\setminus A$ is open and we conclude that $A$ is closed.$\Box$
The interior of $B$ is right and the closure too but there you still need a proof.
But for $C$ you are wrong. You can show $int(C)=C$ and conclude that $C$ is open. This is a bit tricky, but possible.
Proof that $C$ is open:


*

*Let be $(x_0,y_0)\in C$. The euclidean distance to the boundary of
$C$ is $1-x_0^2-y_0^2$. Therefore the euclidean ball with radius of
$1-x_0^2-y_0^2$ and center $(x_0,y_0)$ is still contained in $C$.

*In a euclidean ball with radius $R$ fits a square with edge length $\frac2{\sqrt{n}}R$ where $n$ is the dimension of the space, in this case $n=2$.


This should be known otherwise you have to prove it. 
Now define $\epsilon=\frac{1-x_0^2-y_0^2}{\sqrt{2}}$ and $U:=(x_0-\epsilon,x_0+\epsilon]\times(y_0-\epsilon,y_0+\epsilon]\subset C$. Since $(x_0,y_0)\in U$ we conclude $(x_0,y_0)\in int(C)$. Since $(x_0,y_0)$ was arbitrary we get $C=int(C)$ and $C$ is open.$\Box$
And the closure of $C$ is far too big! 
Claim:
$$
cl(C)=C\cup \{(x,y)~:~x^2+y^2=1\wedge (x>0\vee y>0)\}.
$$
A: The basis you provided is a neighborhood basis.
$A$ is not open because for any $x\in \mathbb{R}$ no neighborhood of $(0,x)$ is included in A.
$Int(A)$ is what you said.
$A$ is closed because all points that do not belong to A have neighborhoods disjont from $A$.
$B$ is open because every point belonging to $B$ has neighborhoods included in $B$.
$B$ is not closed because for any $x \in \mathbb{R}$ every neighborhood of $(x,1)$ has a non-empty intersection and $(x,1) \notin B$.
$Cl(B)$ is what you said.
$C$ is open because every point belonging to $C$ has neighborhoods included in $C$.
$C$ is not closed because $(1,0) \notin C$ and every of its neighborhoods have a non-empty intersection with $C$.
$Cl(C) = C\cup\{(x,y)\in\mathbb{R}^2_+:x^2+y^2=1\}$
Let me know if you need to go into details: proofs are simple but quite verbose.

I will answer here to you comment because it is too long for a comment.
Take a neighborhood of $(x,1)$, that is for fixed $\varepsilon_x>0$ and $\varepsilon_y>0$ use $N=(-\varepsilon_x+x,x]\times (-\varepsilon_y+1,1]$
Now $N\cap B= (-\varepsilon_x+x,x]\times (-\varepsilon_y+1,1] $ if $\varepsilon_y<2$ or $N\cap B= (-\varepsilon_x+x,x]\times (-1,1] $ if $\varepsilon_y\ge 2$.
In either case $N\cap B \ne \emptyset$.
And you know that if all of the neighborhoods of a point has a non-empty intersection with a set it is an adherence point of the set (because no of its neighborhood is included in the complement of the set, that is it is not an interior point of the complement of the set, and you know the the closure of a set is the complement of the interior of the complement of the set)
Let me know if you need further explaination.
