An approach to proving that a set is open. What am I given? What am I trying to show? I'm very new to this sort of mathematics and so I'm a little bit confused with how to approach doing these problems. I look through a lot of example problems and I see them finding the radius to be weird things that just seem to work right after. I do have experience with delta-epsilon proofs but I am struggling much more than these.
Take this example.
Prove that {${(x,y) \in R^2 : x>0}$} is open.
Here's what I've done so far.
Let S represent the set given above. 
S if open if for all P $\in$ S, there exists an $r \gneq o$ such that $B_r(x)$ $\subseteq$ S.
Note that $B_r(x)$ = {$y \in R^2 : d(x,y) < r$}
When I do proofs I usually like to write down my givens and what I need to show.
Given:
Pick a point $p \in$ in S such that $p = (p_x, p_y)$ where $Px \gneq 0$, and then define an open ball $B_r(x)$.
Need To Show:
Pick a point q $\in$ $B_r(x)$, and then show q $\in$ S. 
Is that correct? Because I feel kind of at a loss and not quite sure how this makes much sense... Furthermore I am not quite sure how to go from here either. :/ I keep seeing people just magically come up with r and using the triangle inequality and such to come to a conclusion, but I am not even sure what is being concluded sometimes! How do I come up with this radius? How do I use it to conclude what I want to conclude?
Some serious help appreciated.
 A: The strategy you laid out is okay, although it has a key typographical error. Also, I'm going to re-order things a bit, in order to emphasize that you've missed an important aspect of this proof. 

Given:
Pick a point $p \in S$ such that $p = (p_x, p_y)$ where $p_x > 0$.
Need To Do:
Define an open ball $B_r(p)$. Pick a point $q \in B_r(p)$, and then show q $\in$ S. 

The typographical error was that you had $B_r(x)$ instead of $B_r(p)$.
So, let's carry out this strategy. As written, we pick $p \in S$ such that $p=(p_x,p_y)$ where $p_x > 0$.
Now we have to define an appropriate open ball $B_r(p)$, and to do so you have to define an appropriate radius $r$, and to do that you must write down a formula
$$r = BLAH
$$
where BLAH is an expression using what is given, namely the coordinates $p_x,p_y$.
To be clear, proving the existence of the appropriate radius $r$ by writing a formula for $r$ is the heart of the proof. It is NOT an automatic step. It requires imagination, creativity, and experience.
Now you must ask yourself: What $r$ will work? What $r$ is appropriate? Be imaginative. A picture helps. The set $S$ is the right half plane, and its boundary line is the $y$-axis. The point $p=(p_x,p_y)$ lies to the right of the $y$-axis. Ask yourself: How far to the right? Answer: $p$ lies at distance $p_x$ to the right of the $y$-axis. So, if you pick your radius $r$ to be any positive number less than or equal to $p_x$, then you can probably visualize that the entire open ball around $p$ of radius $r$ stays entirely in the right half plane.
By this imaginative line of thought, you have discovered a precise mathematical formula for $r$ (and one which is amazingly simple):
$$r = p_x
$$
or you may freely choose any positive number less than $p_x$, perhaps 
$$r = p_x/2
$$
suits you better. Either of those is appropriate, and either of them will work for the remainder of the proof.
Armed with that value of $r$, pick a point $q \in B_r(p)$, and then show $q \in S$; that's the easy part.
So to summarize, there's never going to be an automatic proof technique. Whenever you have to prove that some unknown object actually exists and satisfies some desired property (in this case $r$, satisfying the property $B_r(p) \subset S$), the goal is to find that object (the imaginative step), express it precisely (a mathematical description), and then use it in the rest of the proof.
A: let $f:R^2->R$ such that
$f(x,y)=x$.
f is continuous (obvious).
$(0,\infty)$ is an open of $R$
so your set  $f^{-1}((0,\infty))$ is an
open of  $R^2$.
A: It is correct, albeit slightly overcomplicated.  What you want is to show that your $B_{r}(p)$ (not $B_{r}(x)$, as you have it) is a subset of $S$.  Well, to do this, just pick $r$ small enough.  Namely, $r$ should be smaller than the (shortest) distance from $p$ to the boundary of $S$.  This boundary is the line $x=0$.
To help yourself visualize it, sketch the half-plane $S$, then pick, say, $p = (10, -3)$ and see why $r = 9.5$ would work (i.e., would make your $B_{r}(p)$ contained in $S$), but $r = 10.5$ would not.
A: Pick a point $(x,y)\in S=\{(x,y)\in \mathbb{R}^2:x>0\}$. We want to find an open ball centered at the point, which is small enough that it is contained in $S$. Note that all $y$ are permissible, and so are all $x>0$, so we only care about the ball being far enough away from the y-axis. Note that the horizonal distance of $(x,y)$ from the y-axis is just $x$, which is assumed positive since $(x,y)\in S$.
If you take a square centered at $(x,y)$ with side lengths $x/2$ (arbitrary; any length less than $x$ will work), you'll see that the square fits inside $S$ completely. Then take a circle centered at $(x,y)$ with radius small enough that it fits in the square. (You'll see that the square was unnecessary and next time you can just go straight with a small enough circle).
A: ALthough the goal is to understand these at a logical and well-defined level and to get out of the crutch of relying upon "picturing" it in your mind and seeing it is obvious, I think I won't be doing too much harm in "picturing" this.
The set is $S = \{(x,y)\in \mathbb R| x > 0\}$.  So this is the half-plane where everything is to the left of the $x = 0$ axis.  This is "open" because .... it has a fuzzy edge.  Okay, what does that mean? 
We can get as close as we want to the edge without touching it but there are no actual points that or on the edge.  Okay...  that is not math.  We sort of see what they mean but we need a formal way of stating this that is rigorous.
No point is on the edge... that means for any point in $S$ ... we can get closer to the edge.  What do that mean.  What is "the edge"?  Well, the edge is the point that if we go any distance at all we can find a way to step off.   So not being on the edge means... that we can find some distance, maybe a tiny miniscule step, but some distance that no matter what direction we go we won't step off.  
So suppose we are on point $w$ and $w$ is not "on the edge".  Then for some distance $r > 0$ , maybe very  small, we can find all the points within that distance from a point $w$ ... all the points in that distance is $B_r(w) = \{x \in X| d(w,x) < r\}$ ... none of them are "over the edge".  i.e $B_r(w) \subset S$.  That's why the definition works.
And $S$ is open is we can find such a neigborhood for every $w \in S$.
So how do we find such a neighborhood of $w$?  By finding a really small distance $r$.  And how do we find that really small distance?
Well, this is the big secret no-one discusses.  The small distance, $r$ should be the minimum distance $w$ is from the edge.  (Okay, not always.  Sometimes, particularly if you have many points to consider you need to find half the distance so that values don't "compound")
So let $w = (a,b) \in S = \{(x,y)| x > 0\}$.   So $a > 0$ but $b$ could be anything.  So $w$is not "on the edge".  It's ... some distance ... from the edge.  What distance?   Well the edge is $\{(0,y)\}$ so it's ... $a$ away from the edge.
$B_a(a,b) \subset S$??? That's what we have to prove?  Okay.  ... um, what are we proving again?  $B_a(a,b) \subset S$.  That means... if $(c,d) \in B_a(a,b)$ then $(c,d) \in S$.  
$(c,d) \in B_a(a,b)$ means ... what?  ... $d((c,d),(a,b)) = \sqrt{(c-a)^2 + (b-d)^2) } < a$.  
$(c,d) \in S$ means.... $c > 0$.
So $\sqrt{(c-a)^2 + (b-d)^2) } < a$
$(c-a)^2 + (b-d)^2 < a^2$
$(c-a)^2 < a^2 - (b-d)^2$ and ... umm... where are we going with this...
$|c - a| < \sqrt{a^2 -(b-d)^2}$.  How on earth are we going to get $c > 0$.... 
Well, if $c \ge a$ then... $c \ge a > 0$.  So we can ignore that case.
We just need to worry if $c < a$.
$a - c < \sqrt{a^2 - (b-d)^2}$  so 
$c > a - \sqrt{a^2 - (b-d)^2}$ and .... um?  ... dag!  Does $a - \sqrt{a^2 - (b-d)^2} > 0$?
well, yeah.. $(b-d)^2 \ge 0$ so $a^2 - (b-d)^2 < a^2$.  so $\sqrt{a^2 - (b-d)^2} \le \sqrt{a}$.... wait.  How do we know $a^2 - (b-d)^2 > 0$?  
Well, earlier we had $(c-a)^2 < a^2 - (b-d)^2$ so $0 < (c-a)^2 < a^2 - (b-d)^2 <  a^2$
So $a - c < \sqrt{a^2} = |a| = a$.
So $c > a - a = 0$.  So..... phew....
$(c,d) \in S$ so $B_a((a,b)) \subset S$ so $(a,b)$ is an interior point of $S$, and $(a,b) $ was arbitrarily picked, so all points of $S$ are interior points, so $S$ is open, which "means" no point $(a,b)$ is actually on "the edge" so $S$ has a "fuzzy border".
Phew.  That's how I figured it all out.  It gets more intuitive the more you practice.
A: A set $U$ in a metric space is open if for every $p\in U$ there exists $r_p>0$ such that $B(p,r_p)\subset U.$ Because $\cup_{p\in U}B(p,r_p)$ is open , and $$U=\cup_{p\in U}\{p\}\subset \cup_{p\in U}B(p,r_p)\subset U.$$ In a specific case we may have to choose  a suitable $r_p$ for each $p\in U.$
In your Q, if $p=(x,y)$ with $x>0,$ and $r_p=x/2,$ observe that if $p'=(x',y')$ with $x'\leq x/2$ then the distance $d(p,p')=\sqrt {(x-x')^2+(y-y')^2}$ is at least equal to $|x-x'| ,$ which is at least  $x/2,$ implying  $(x',y')\not \in B(p,x/2).$.... So if $(x'',y'')\in B(p, r_p)=B(p,x/2)$ then $x''>x/2>0.$... It may help to draw a diagram.
