$A \subseteq \mathbb{R}^n$ is closed iff for any $\{x_i\}_{i=1}^{\infty}$ in $A$ with limit $x$, we have that $x \in A$ PROBLEM:
$A \subseteq \mathbb{R}^n$ is closed iff for any $\{x_i\}_{i=1}^{\infty}$ in $A$ with limit $x$, we have that $x \in A$.
MY SOLUTION:
Say $A$ is closed, and pick a sequence $\{x_i\}_{i=1}^{\infty}$ of $A$ such that $\lim(x_i) = x$. We want to show $x \in A$. If $x \notin A$, then there exists an $r>0$ such that $D^n(x,r) \cap A = \varnothing$. But, we know $\lim(x_i) = x$, therefore there exists $j>0$ such that $||x_i - x|| < r$ for $i \geq j$. In other words, $x_i \in D^n(x,r)$. But, we are given $x_i \in A$. Therefore, $D^n(x,r)$ and $A$ must overlap. Contradiction. Hence $x \in A$.
Conversely, suppose for any $\{x_i\}_{i=1}^{\infty}$ in $A$ with limit $x$, we have that $x \in A$, we show $A$ is closed. Pick a point $x \notin A$, so we want to find an $r>0$ such that $D^n(x,r) \cap A = \varnothing$. Suppose this is false and put $r = \frac{1}{N}$. Therefore, we can find an $x_N \in D^n(x,\frac{1}{N}) \cap A$. Hence, $||x_N - x|| < \frac{1}{N}$. This implies $\lim(x_N) = x$. But, we assumed $x_N \in A$, therefore, $x \in A$ which is a contradiction. Therefore, $D^n(x,r) \cap A = \varnothing$ and by definition $A$ must be closed.
Is this solution correct? any feedback? thanks
 A: The first half is okay, and you have the right idea for the second half, but the second half is worded badly enough to be quite confusing. Let me give a corrected version and then try to explain what’s wrong with what you wrote. (Some of my changes are merely cosmetic: for instance, I don’t like using upper-case letters for integers, so I’ve replaced your $N$ with $k$.)

Conversely, suppose that whenever a sequence in $A$ converges to some point $x\in X$, then $x\in A$; we want to show that $A$ is closed. Pick a point $x\in \Bbb R^n\setminus A$; we want to show that there is an $r>0$ such that $D^n(x,r)\cap A=\varnothing$. Suppose that this is false; then for each $k\in\Bbb Z^+$ there is a point $x_k\in D^n\left(x,\frac1k\right)$. Thus, $\|x_k-x\|<\frac1k$ for each $k\in\Bbb Z^+$, so the sequence $\langle x_k:k\in\Bbb Z^+\rangle$ converges to $x$, and by hypothesis $x\in A$. This is a contradiction, so for each $x\in\Bbb R^n\setminus S$ there is an $r_x>0$ such that $D^n(x,r_x)\cap A=\varnothing$, and $A$ is therefore closed.

You ‘put $r=\frac1N$’ after assuming that there is no $r>0$ such that $D^n(x,r)\cap A=\varnothing$, but this doesn’t actually make sense as written: it says that you already have some $N$, and you’re defining $r$ to be the reciprocal of that $N$, which isn’t the case. What you really meant is that for each $k\in\Bbb Z^+$ you wanted to apply the assumption that $D^n(x,r)\cap A\ne\varnothing$ with $r=\frac1k$ in order to conclude that there is an $x_k\in D^n\left(x,\frac1k\right)\cap A$. You need to make it clear that you’re actually choosing a sequence of points; as you wrote it, at best it appears that you’re choosing one $x_N$.
