Determine whether $M = \{(x, y) \ | \ y < \sin(\frac{1}{x}) \ \text{and} \ x > 0 \} \subseteq (\mathbb{R}^2, d)$ is open or closed or neither 
Determine whether $M = \{(x, y) \ | \ y < \sin(\frac{1}{x}) \ \text{and} \ x > 0 \} \subseteq (\mathbb{R}^2, d)$ is open or closed or neither

Note that $M \subseteq (\mathbb{R}^2, d)$, where $d$ is the usual metric on $\mathbb{R}^2$, hence $M$ is a metric space and open or closed here is relative to the metric topology on $\mathbb{R}^2$
My Attempted Proof: 
Fix $y \in \mathbb{R}$ such that $y < -1$ and let $(\alpha_n)_{n \in \mathbb{N}}$ be defined such that $\alpha_n = (\frac{1}{n}, y)$, then $\lim_{n \to \infty} \alpha_n = (0, y) \not\in M$, and since in metric spaces, sequential limits coincide with topological limits $M$ cannot be closed since it doesn't contain its limit points.
To show $M$ is not open, we show that $\mathbb{R}^2  \setminus M$ is closed. We have $$\mathbb{R}^2  \setminus M = \left\{(x, y) \ | \ y \geq \sin\left(\frac{1}{x}\right) \ \text{and} \ x\leq 0 \right\}$$. We claim that $\text{Bd}(\mathbb{R}^2 \setminus M) = \{ (0, y) \ | \ y \in [-1, 1]\}\  \cup \ \{(x, \sin(\frac{1}{x})) \ | x < 0 \} $. To prove this claim, pick $p = (0, y)$ for some $y \in [-1, 1]$, $p$ is a boundary point since for any neighbourhood $U$ of $p$ there exists an open ball $B = B_{(\mathbb{R}^2, d)}(p, \epsilon) \subseteq U$ for some $\epsilon > 0$ such that $B$ intersects $\mathbb{R}^2 \setminus M$ in at least one point  and it intersects $M$ in at least one point ((based on the wild oscillations of $\sin(\frac{1}{x})$ near $0$). ---------    (*) 
Now pick $q = (x, \sin\left(\frac{1}{x}\right))$ for some $x < 0$, then trivially any neighbourhood of $q$ intersects bot $M$ and  $\mathbb{R}^2 \setminus M$ in at least one point, and for any other point $p' \in \mathbb{R}^2 \setminus M$ we can always find an open ball around $p'$ small enough that it only intersects $\mathbb{R}^2 \setminus M$. Hence $\text{Bd}(\mathbb{R}^2 \setminus M) = \{ (0, y) \ | \ y \in [-1, 1]\}\  \cup \ \{(x, \sin(\frac{1}{x})) \ | x < 0 \} $
However $p \not\in \mathbb{R}^2 \setminus M$ as $\sin(\frac{1}{x})$ isn't even defined at $x = 0$. So $\text{Bd}(\mathbb{R}^2 \setminus M) \not \subseteq \mathbb{R}^2 \setminus M$. Hence $\mathbb{R}^2 \setminus M$ is not closed, and $M$ is not open. $\square$

Firstly is this proof correct and rigorous and statisfactory enough? 
Secondly, I'm not too happy about the (*) part, because at best all I'm doing is just taking an educated guess based on the plot of $\sin(\frac{1}{x})$. Is there any other way I can make my proof of my claim of the boundary more rigorous? Also is there any other way I can go about proving that $M$ is not open (or that $M$ is open if I'm incorrect)?
 A: Your proof that $M$ is not closed went very well. Unfortunately, your other portion has a glaring issue: you've mis-determined the complement of $M.$ Rather, it should be $$\left\{(x, y)\in\Bbb R^2 : \ y \geq \sin\left(\frac{1}{x}\right) \ \text{or} \ x\leq 0 \right\}.$$ Also, it turns out that the boundary of $M$ is instead $$\left\{(0,y)\in\Bbb R^2:y\le 1\right\}\cup
\left\{\left(x, \sin\left(\frac{1}{x}\right)\right): x>0 \right\},$$ which can be shown to lie in the complement of $M,$ whence $M$ is, indeed, open. In particular, note that if $p\in M,$ then the $x$-coordinate of $p$ is positive. Thus, by picking $p=(0,y),$ you've guaranteed that $p\notin M,$ regardless of what $y$ is.
Another, easier way to go (if you have the results at hand to implement it) is to use the fact that continuous pre-images of open sets are open. Then, show that $f(x,y):=\sin\left(\frac1x\right)-y$ is a continuous, real-valued function on the open set $\{(x,y)\in\Bbb R^2:x>0\},$ and that $M$ is precisely the preimage of the open set $(0,\infty)$ under $f,$ so is open in $\{(x,y)\in\Bbb R^2:x>0\},$ when the latter is considered as a subspace of $\Bbb R^2.$ Since open subsets of open subspaces are open in the overlying space, then $M$ is open in $\Bbb R^2,$ as desired.

Added: An up-side, here, is that since $M$ is a non-empty open proper subset of the euclidean space $\Bbb R^2,$ then you no longer have to prove that $M$ isn't closed.
