Show the set is open I've been seriously trying to solve this for really lots of time. It's pretty basic, which furthers the frustration... 

Let there be a set $A=\{x=(x_1,x_2)\in \mathbb R^2 :|x_1|+|x_2|<1\}$. Show it is an open set. 

So first of all I say it's all the points within a square in the plane on $(1,0), (0,-1), (-1,0)$ and $(0,1)$. 
I took the following approach - and if you could, I would like you to show me or hint me how to solve in this approach: Let there be $P = (x_1,x_2) \in A$, and lets show $P$ is an interior point. 
Now that's what I got stuck with. I want to define a circle $B((x_1,x_2),r)$ for some radius $r$, and finding $r$ is what I got problems with. 
My attempt: Lets assume $|x_1| \leq |x_2|$, and so we'll choose $r=1-|x_2|$, and define $B((x_1,x_2),r)$. And lets take $(y_1,y_2) \in B$, and I need to prove that $(y_1,y_2)\in A$. Just couldn't make it...
Thanks for any help in advance! 
 A: You can do it geometrically. Take a point in your set, and suppose it is in the first quadrant. Drop lines parallel to the axes to the line $x+y=1$, and mark half the distance to the intersection. Then the circle with that radius is entirely contained in the square, because the diagonal is strictly longer that the radius.

A: There are several ways to show it:
Let the function $f: \mathbb R^2\to\mathbb R$, with
$$
f(x_1,x_2)=\lvert x_1\rvert+\lvert x_2\rvert.
$$
Clearly $f$ is continuous and as $(-\infty,1)$ is open in $\mathbb R$, so is its inverse image:
$$
f^{-1}(-\infty,1)=\big\{(x_1,x_2)\in\mathbb R^2: \lvert x_1\rvert+\lvert x_2\rvert<1 \big\}.
$$
A: As you said let $P=(x_1,x_2)$ then set $r_0=|x_1|+|x_2|$ then it is on the square $|x|+|y|=r_0$ and $r_0<1$.
Then choose $r=(1-r_0)/\sqrt 2$ then $B(P,r)$ is insede of the big square. 
Why did I choose $r$ in that pattern, it is completley geometric,Just drow the squares!
A: The complement of $A$ contains its boundary. Hence, $A^C$ is closed, so $A$ is open.
