Rudin Exercise 2.8 I was hoping someone could read over and critique my proposed proof of Rudin exercise 2.8. 

Is every point of every open set $E \subset \mathbb{R}^2$ a limit point of $E$? Answer the same question for closed sets in $\mathbb{R}^2$. 

Here is what I've been able to put together. 

Let $E \subset \mathbb{R}^2$ be an open subset of $\mathbb{R}^2$. Let $x \in E$. Since $E$ is open, there exists a neighborhood $N_{r} (x) \subset E$ where $r > 0$ and
  $$
N_r (x) = \{y \in E \mid d(x,y) < r \}.
$$
  To show that $x$ is a limit point of $E$, it suffices to show that any neighborhood of $x$ contains some point in $E$ other than $x$. Let $s > 0$, and consider the neighborhood $N_s (x)$, where
  $$
N_s (x) = \{y \in E \mid d(x,y) < s\}.
$$
As $x \in \mathbb{R}^2$, we can write $x = (a,b)$. Now, let us construct a new point $y \in \mathbb{R}^2$, where 
  $$
y = \left(a + \frac{1}{2}\min(r,s), b\right). 
$$
  Due to the clear disagreement in the first component, these are distinct points.
Thus, we have: 
  \begin{align*}
d(x,y)  & = d\left((a,b), \left(a + \frac{1}{2} \min(r,s), b\right)     \right) \\
& = \sqrt{\left(a - \left(a + \frac{1}{2} \min(r,s)\right) \right)^2 + (b-b)^2} \\
& = \sqrt{\left(0 - \frac{1}{2} \min(r,s)\right)^2 + 0} \\
& = \sqrt{\frac{1}{4} \left(\min(r,s)\right)^2} \\
& = \left \lvert \frac{1}{2} \min(r,s) \right \rvert \\
& = \frac{1}{2} \min(r,s) \\
& \leq \frac{1}{2} \cdot r \\
& < r.
\end{align*}
  Thus, $d(x,y) < r$, so $y \in N_r (x) \subset E$, and hence $y \in E$. Furthermore, by virtually exactly the same arithmetic, we have $y \in N_s (x)$: 
  \begin{align*}
d(x,y) & = |x-y| \\
& = d\left((a,b), \left(a + \frac{1}{2} \min(r,s), b\right)     \right) \\
& = \left \lvert \frac{1}{2} \min(r,s) \right \rvert \\
& = \frac{1}{2} \min(r,s) \\
& \leq \frac{1}{2} \cdot s \\
& < s
\end{align*}
  So $d(x,y) < s$, and $y \in N_s (x)$. Since $s > 0$ is arbitrary, we see that given an $s > 0$, we can find a point $y$ such that $y \in N_s (x)$ and $y \in E$, so $x$ is a limit point of $E$. Since $x$ is an arbitrary point of $E$, we conclude that every point of $E$ is a limit point of $E$. 
If $E$ is replaced with a closet set, it is not the case that every point of $E$ is a limit point. As a counterexample, consider any finite set, which we know to be closed. However, finite sets have no limits. 

How does this proof look? The biggest question I have at this point is whether it is too far of a jump to assume the Euclidean metric on $\mathbb{R}^2$. It seems to me that we can define a metric on $\mathbb{R}^2$ with a different metric. However, I do not know how to produce the same results, $d(x,y) < r$ and $d(x,y) < s$ with an "arbitrary distance function." Can I assume the Euclidean metric? Are there theorems on the subject that I've missed? Perhaps there's some use here for the triangle inequality to produce those same results. 
Any help would be greatly appreciated.
 A: Validity of your proof
Your proof looks fine. You can assume the usual Euclidean metric given that you're told that you're working in $\mathbb{R}^2$ and Rudin is looking at the usual metric on $\mathbb{R}^2$. 
Comments on Proof structure
Here is what I will say about the proof in terms of writing and I'll answer some of your questions along the way. Your proof is good in terms of the logical reasoning - there is no hand-waving or leaps in logic that are left in the air, and each statement very cleanly follows from the preceding one. This is often one of the best skills to have an intuition for when you start writing proofs and you seem to be more than fine there. 
Now a good thing to keep in mind is that it's always helpful to take a step back and outline the basic steps of the idea. Here is essentially the sketch (and, at least as I believe, the intuition) behind what you're saying:
Step 1: Consider a point x in $\mathbb{R}^2$ and an open ball/neighborhood around x with some point in E.
Step 2: Take a smaller open ball/neighborhood around x 
Step 3: Show that the smaller open ball has a point y such that y is in E.
Notice that the central intuition that you picked up on that works here is that once you get one open ball around x that is fully contained in E, we can keep shrinking it to a smaller ball that contains x and still is a subset of E (and so contains your y). While it's good that you worked through and constructed a specific y, writing the proof for a case where we're working in higher dimensions (say, $\mathbb{R}^{57}$) seems like it would get more tedious, since the metric involves more and more variables. But that should strike you as odd, since the central idea does not change. We can rephrase your proof as follows:
Suggested Refinement
Let x $\in$ E and let N$_r$(x) $\subset E$ be an open ball of radius r $>$ 0 centered at x. We will call this set N$_1$. Pick an arbitrary point $x_1$ in N$_1$ that is not equal to x. 
Now consider N$_2$ given by N$_2 = $ N$_{r/2}$ the open ball of radius r/2 $ > 0$ centered at x. Clearly N$_2 \subset $ N$_1$. Pick a point $x_2$ in N$_2$ not equal to x. 
In general, let N$_n = $ N$_{r/n}$ and let $x_n \neq x \in N_n$. Then by definition, the limit of {$x_n$} = x. 
Of course, with a little more effort, you can make certain statements here a little more rigorous, but the idea is that this is essentially what is happening in the proof. But this can now be generalized to any $\mathbb{R}^n$ because we never need to explicitly use the distance formula. This gives us another useful insight to one of your questions. 
What this tells you about the larger idea
The key idea that's at play is that you can always find a point in your smaller neighborhood that is distinct to the original point x. It's the reason this idea works. So to answer your question whether you can define any metric on $\mathbb{R}$ and have this still work - the answer is no. The easiest way to see this is the trivial metric given by d(x,y) = 0 if x = y and 1 if not. See that under this metric, not only does the proof fail but the statement itself is false. So while you can prove the same theorem under a slightly more general case of the standard metric (as shown above), you can't do it for any arbitrary metric. 
In Conclusion
You seem to be doing more than well with the kinds of questions you're asking and the rigorous approach to thinking about problems. Something to keep in mind is along with what you're doing, take a step back and see what the structure or sketch of your argument is. Ask yourself what the essential moving parts / ideas are, which hypotheses you use when, and what would happen if you alter / remove them. Then you'll see more and more of what these parts tell you about the theory as a whole and you'll be able to play with it. It's something you'll get better with the more you practice. But that's a part of the fun!
