Let $T = \{x\in\mathbb{R}: 2 x^2 \cos(1/x) \le 1 \} \cup \{0\}$. Is this is a complete set in $\mathbb{R}$ with respect to usual metric?
I am thinking about using the fact that the inverse image of a closed set under a continuous function is closed, but here $2 x^2 \cos(1/x)$ is continuous on $\mathbb{R}/\{0\}$, since it's not defined at $x=0$. On the set $A=[-\infty , 1]$ which is closed , $f(x)$ is not continuous (due to zero). On $P=\{x\in\mathbb{R}\;\mathrm{such\;that\;} 2 x^2 \cos(1/x) >1\}$, $2 x^2 \cos(1/x)$ is continuous , and $(1,\infty)$ is an open set so $P$ is an open set. Now the complement of $P$ is $K=\{x \in \mathbb{R} \;\mathrm{such\;that}\; 2x^2 \cos(1/x) \leq 1\}$ is a closed set (?) and $\{0\}$ is a closed set so $T\setminus K \cup\{0\}$ is a closed set. Am I right? Sorry for my english.
You're on the right track, judging from the comment you made, but one of the sets you claim to be closed isn't (it's what you call $K$). No big deal, anyway: it can be fixed.
Consider the function $$ f(x)=\begin{cases} 2x^2\cos\dfrac{1}{x} & x\ne0 \\[6px] 0 & x=0 \end{cases} $$ This function is clearly continuous at every $x\ne0$; on the other hand, $$ \lim_{x\to0}f(x)=0 $$ because you can squeeze $$ -2x^2\le2x^2\cos\frac{1}{x}\le 2x^2 $$ (for $x\ne0$).
Now $T=\{x\in\mathbb{R}:f(x)\le 1\}$, so…