Show that subset $T = \{(x,y) \mid x \neq 0, -3I let $X=(x_1, y_1)$ be a point in $T$, let $B(r,X)$ be the ball of radius $r$ centered at $X$, and let $Y = (x_2,y_2)$ be a point in $B(r,X)$.
So i want to show that $Y$ is in $T$, in other words that $x_2$ not equal to $0$ and that $-3<y_2<3$.
I started with $\sqrt{(y_2-y_1)^2 + (x_2-x_1)^2} < r$
...
$|y_2 - y_1| < r$  and  $|x_2 - x_1| < r$
...
$y_1-r < y_2 < y_1+r$  and  $x_1-r < x2 < x_1+r$
so I set $y_1 - r = -3$  and $y_1 + r = 3$
which led me assuming $r$ must be equal to the $\min(y_1 + 3, 3 - y_1)$
My questions are the following:
How do I use this to show that $Y$ is in $T$ ?
What do I do with the fact that $X$ cannot equal $0$?
Is my reasoning correct?
 A: Your posting is a bit confusing.
I assume you want construct an $r \in \Bbb R$ s.t. for all $Y \in B(r,X)$ also $Y \in T$ holds?
Your first conclusion is already not useful.
You wrote:

I started with $\sqrt{(y_2-y_1)^2 + (x_2-x_1)^2} < r$
...
$|y_2−y_1|<r$ and $|x_2−x_1|<r$

If you mean an implication by "..." this is indeed true.
So let's discuss what we actually want and what we have:
You ḱnow: $X = (x_1,y_1) \in T$ hence $x_1 \not= 0, -3 < y_1 < 3$
We want: $Y=(x_2,y_2) \in T$ hence $x_2 \not= 0, -3 < y_2 < 3$
So: How to choose $r\in \Bbb R$ s.t. the following implication holds:
$|y_2−y_1|<r \wedge|x_2−x_1|<r \quad \Rightarrow \quad x_2 \not= 0, -3 < y_2 < 3$
1.) Let's ensure $x_2 \not= 0$ first. We know $x_1 \not= 0$ so if $r < |x_1|$ it follows directly $x_2 \not = 0$.
2.) To ensure $-3 < y_2 < 3$ we have choose $r < 3 - y_1$ as well as $r < y_1 + 3$ (this is what you already got as I realize now ^^)
So to ensure $Y \in T$ we have to choose $r \in \Bbb R$ this way it fulfills both 1.) and 2.)
So choosing $r < \min\{|x_1|, y_1 + 3,3 - y_1\}$ gives us the wanted result.
