Let us look at $\mathbb{R}^2$ with the Euclidean norm. Is the set $\{\left(x,\sin\frac{1}{x}\right) | x\in(0,\infty)\}$ with the induced metric from $\mathbb{R}^2$ a complete space?
I think it is, since $f(x) = \sin\frac{1}{x}$ is continuous in $(0,\infty)$, and therefore the set is closed, and we know every closed subset of a complete metric space is complete with the induced metric. This seems ok, but for some reason since $\sin\frac{1}{x}$ is somewhat of a irregular function, I have some doubts. Is this true? Is the proof OK?
Edit: I will add another way I thought to show this: Suppose $(x_n,y_n)\in\{\left(x,\sin\frac{1}{x}\right) | x\in(0,\infty)\}$ is a Cauchy sequence. This means that for all $\varepsilon>0$, there exists $N \in\mathbb{N}$ s.t. for all $n,m>N$, $\sqrt{(x_n-x_m)^2 + (y_n-y_m)^2} < \varepsilon$. By definition, $y_n = \sin\frac{1}{x_n}$. Now, since $(x_n,y_n)$ is Cauchy, then each coordinate is Cauchy in $\mathbb{R}$, hence $\sin\left(\frac{1}{x_n}\right)$ is Cauchy in $\mathbb{R}$. But since $\mathbb{R}$ is complete, this means that there exists $a\in\mathbb{R}$ such that $\sin\left(\frac{1}{x_n}\right)\to a$. Since $\sin\left(\frac{1}{x}\right)$ has no limit at $x = 0$, we conclude that $\frac{1}{x_n}\not\to 0$. Since $x_n$ is Cauchy, we conclude that $x_n\to b>0\in \mathbb{R}$. Overall, we get that $(x_n,\sin\left(\frac{1}{x_n}\right))\to (a,b)$ and since $\sin\left(\frac{1}{x}\right)$ is continuous at $x>0$, we get that $b = \sin\left(\frac{1}{a}\right)$, and therefore $(a,b)\in\{\left(x,\sin\frac{1}{x}\right) | x\in(0,\infty)\}$.
Does this hold?
I believe that space isn't complete:
Let $A = \{x\in(0,\infty)$ | $sin(\frac{1}{x}) = 1\}$
$A$ is an infinite set, so let $\{x_n\} \subset A$ be a non repeating sequence s.t $x_n \to 0$
Hence $\{(x_n,1)\}_{n\in \mathbb{N}} \subset \{\left(x,\sin\frac{1}{x}\right) | x\in(0,\infty)\}$
But $\{(x_n,1)\}_{n\in \mathbb{N}}$ is convergent and thus cauchy, with a limit not in the graph.