proving that $D = \{(x,y) \in \mathbb R ^2: x^2 + y^2 < 1\}$ is opened In a general topology exercise I have to prove the following:

Prove that the disk $D = \{(x,y) \in \mathbb R ^2: x^2 + y^2 < 1\}$ is opened in the euclidean topology.

This reminded me in how in multi-variable calculus we approximates the regions in the plane with little rectangles in order to integrate over that region. But I have no idea how to approach the problem. How should I prove this? Any tips?
Edit: This exercise is in the beginning of the book, At this point the only concepts that I'm allowed to use in the proof are the following definitions and concepts:

*

*Definition of topology on a set

*Definition of basis of a topology on a set

*The basis for the euclidean topology is $B=\{(x,y)\in \mathbb R ^2:a < x < b \wedge c < y < d \}$
I have not yet reached the chapter about metric spaces, so I'm not allowed do define the disk as an opened ball using some metric $d$. I think that the objective of the exercise is to prove that it is opened using the basis of the euclidean topology.
 A: You can use the more general fact that, if $V\subset\mathbb{R}$ is an open set and $f:\mathbb{R}^2\rightarrow\mathbb{R}$ is a continuous function, then $f^{-1}(V)=\{x\in \mathbb{R}^2:f(x)\in V\}$ is an open set in $\mathbb{R}^2$. In your case, take $V=(-\infty,1)$.
Edit: a more direct approach is as follows. Let $D$ be the open unit disk and $x\in D$. If we take $\delta=1-|x|>0$, where $|x|$ denotes the norm, then you can check that the open disk centered at $x$ with radius $\delta>0$ is completely contained in $D$, i.e.
$$B_{\delta}(x)\overset{\text{def}}{=}\{y\in\mathbb{R}^2:|x-y|<\delta\}\subset D.$$
Then you draw a suitable open rectangle (that's part of the basis) inside $B_{\delta}(x)$, let's call it $R_x$. Then observe that
$$D=\bigcup_{x\in D}R_x.$$
Therefore, $D$ is an union of basis sets, so $D$ is open.
A: Given a point p $\in$ D:
Let M = $1 - d(p,0)$, then the open set $B_M (p)$ = {q $\in \mathbb{R}^2 :$ d(p,q)< M} is contained in D. Where d is the euclidean metric.
Note that: D is the Union of an arbitrary number of these open balls.
Conclusion, D is open.
A: Here I propose another way to solve it for the sake of curiosity.
Let's consider that $x\in B(x_{0},R) = \{z\in\textbf{R}^{n}\mid d(z,x_{0}) < R\}$. We shall prove the ball $B(x,r)$ is contained in $B(x_{0},R)$, where $r = R - d(x,x_{0})$. Indeed, one has that
\begin{align*}
p\in B(x,r) \Rightarrow d(p,x_{0}) & \leq d(p,x) + d(x,x_{0}) < r + d(x,x_{0})\\\\
& = (R - d(x,x_{0})) + d(x,x_{0}) = R \Rightarrow p\in B(x_{0},R)
\end{align*}
and we are done. Hopefully this helps.
