Why does paracompactness need to be not locally finite open subcollection, but refinement of opem covering? In Munkres topology, he defines paracompactness as a generalization of compactness, as follows:

$X$ is paracompact if every open cover of $X$ has locally finite open refinement that covers $X$

But why it should be "refinement" of open cover? In compact, we need "subcollection". Surely subcollection is more strong assumption, but then , is there any paracompact space that has refinement but not subcollection?
If then, what important difference between refinement and subcollection does make difference of definition in paracompactness?
 A: Take the open interval $(0,1)$ and consider its open sets
$$U_n=(1/n,1-1/n)\qquad n\geq2.$$
Then $\mathcal{U}=\{U_n\}_{n\geq2}$ is a countable open cover of $(0,1)$ which has no locally-finite subcover. To see this observe the stronger fact that any locally-finite subcover of $\mathcal{U}$ is finite, and $\mathcal{U}$ has no finite subcover.
Of course $(0,1)$ a separable, locally compact metric space homeomorphic to $\mathbb{R}$. Evidently it is paracompact and $\mathcal{U}$ has a locally-finite refinement.
Note that in general a space $X$ is compact if and only if every open cover of it has a finite subcover if and only if every open cover of it has a finite refinement. If you consider this second condition to be the definition of compactness, then paracompactness is its logical extension.
A: The property "every open covering has a locally-finite subcovering" is equivalent to compactness. So any paracompact space that is not compact fails to have this property.
In fact, even the property of having point-finite subcoverings is equivalent to compactness. (Pick $U_0 \in \mathcal{U}$ and take a point-finite subcovering of $\{U \cup U_0 | U \in \mathcal{U} \}$.)
