Let $M = \{(x,y) \in \mathbb{R}^2 : y = \sin{\frac{1}{x}}\}$ be a set with a Euclidean distance metric. I need to determine whether $M$ is complete.
I suppose
if M is complete we need to prove the following implication: $(\forall E > 0 \ \ \exists N \in \mathbb{N}: \forall n,m > N \ \ \sqrt{(x_n-x_m)^2 + (\sin{\frac{1}{x_n}} - \sin{\frac{1}{x_m}})^2} < E) \Rightarrow (\exists A = (A_1,A_2) \in \mathbb{R}^2 :\forall E > 0 \ \ \exists N \in \mathbb{N}: \forall n > N \ \ \sqrt{(x_n-A_1)^2-(\sin{\frac{1}{x_n}}-A_2)} < E)$
if M is not complete, we need to find a counterexample. Since $\mathbb{R}$ is complete, a potential counterexample would be a sequence of points of $M$ whose limit will be in $\mathbb{R}^2 \setminus M$. That said, every fundamental sequence of points $(x_n, \sin{\frac{1}{x_n}})$ has a limit $A = (A_1,A_2)$ in $\mathbb{R}^2$. If $M$ in incomplete, then $A_2 \neq \sin{\frac{1}{A_1}}$.
$(0,0)$ is a limit point but $(0,0)\notin M$ (Put $x_n=\frac{1}{2n\pi} \Rightarrow x_n\to 0 , \sin x_n=\sin \frac{1}{\frac{1}{2n\pi}}=\sin 2n\pi=0$
Theorem Subspace of Complete Metric Space is Closed if and only if Complete.