Bolzano-Weierstrass property and compactness

Question

In my notes of functional analisys there is this definition of compact set in a metric space: $K$ is compact if and only if $\forall$ bounded sequences $(f_k)_k\subset K$ there exist a convergent subsequence $(f_{k_n})_n\subset (f_k)_k$ to an element of $K$.

In other word a set $K$ is compact if and only if it verify the Bolzano-Weierstrass property. But if we consider $\mathbf{R}$, we know that it verify the B-W property, so the definition above tell me that $\mathbf{R}$ is a compact space. (!Contraddiction!)

What is the problem? Does the definition above has some tacit property? In general topology I know that a set $K$ is compact if and only if all its sequences have a convergent subsequences in $K$, and obviously, this definition is not verifyed in $\mathbf{R}.$

Answer 1

Because a subset $K$ of a metric space $X$ is compact if and only if every sequence of elements of $K$ has a subsequence which converges to an element of $K$, not just the bounded ones.

Answer 2

Thank you. Do exist some cases where all the sequences in a metric space are bounded? Obviously when the space is compact, but is this the only one case?