Example of a metric space where Heine-Borel theorem does not hold In Rudin, it says the Heine-Borel theorem holds for Euclidean metric spaces.
What is an example of a metric space where Heine-Borel does not hold true?
 A: Any non-complete metric space, or an infinite-dimensional Banach space.
A: Consider $\Bbb R^2 \setminus \{(0,0)\}$ with the usual metric restricted from $\Bbb R^2$. The set $$D = \{ (x,y) \in \Bbb R^2 \mid 0 < x^2+y^2 \leq 1 \}$$is closed in $\Bbb R^2 \setminus \{(0,0)\}$, bounded, but not compact. Sequences in $D$ which "want" to converge to $(0,0)$ don't have limit in $D$.
A: Motivated by a comment on another answer, here goes another example (which avoids the psychologically problematical "non-complete" case, while being elementary):
Let $(M,d)$ be an infinite set with the discrete metric. $M$ is then bounded, obviously closed as a subset of itself, but is not compact (the cover $\{x\}_{x \in M}$ is an open cover with no finite subcover). Note that $M$ is also complete.
A: Consider the space 
$$
c_0=\{(x_n)_n:\ \lim_nx_n=0\},
$$
with the distance $$d(x,y)=\sup\{|x_n-y_n|:\ n\in\mathbb N\}.$$
The unit ball $$B=\{x\in c_0:\ d(x,0)\leq1\}$$ is closed, bounded, and it contains the sequence $$E=\{e^k:\ k\in\mathbb N\},$$ where $e^k$ is the sequence with entries $\delta_{k,n}$, i.e., $e^k$ consists of all zeroes and a one in the $k^{\rm th}$ position. It is easy to see that $$d(e^k,e^h)=1,\ \ h\ne k,$$ so the sequence $E$ admits no convergent subsequence. 
