Boundedness and total boundedness
Solution 1:
Hint: Consider an infinite set in the discrete metric. (Every point is at distance $1$ to every other point.)
Solution 2:
Consider the set of real numbers $\mathbb{R}$ with the discrete metric defined as
$$ d(x,y)= \begin{cases} 1,&\quad x\ne y \\ 0,&\quad x=y \end{cases}. $$
Then every open ball with this metric is
$$B_{\mathbb{R}}(x,r)= \{y\,|\,y\in\mathbb{R}\,\land d(x,y)<r\}= \begin{cases} \{x\},& \quad r\leq 1 \\ \mathbb{R},& \quad r>1 \end{cases}. $$ Accordingly, every subset of $\mathbb{R}$ is closed, open and bounded with the discrete metric, including the $\mathbb{R}$ itself! But clearly $\mathbb{R}$ is not totally bounded for if we choose $\epsilon = \frac{1}{2}$ then there won't be any finite covering for it!