To prove compactness of some subset Question
Show that (0,1] is not compact by constructing an open cover (0,1] that does not have a finite sub over
-> since set is compact if and only if every sub open cover is an finite set
But open cover (0,1] have infinite sub open cover such as (0,1-1/n) n=0 to infinity
So (0,1] is not satisfy the compactness 
Is this reasonable proof?
oh no i mean (0,1-1/n] 

 A: Try $$(0,1] = \bigcup_{n\ge 1} (1/n, 1].$$
A: The set $\{ (a,1] : 0<a\le 1 \}$ is an open cover of $(0,1]$ and there is no finite subcover.
A: I think some of the existing answers are going to lead to some confusion. Note that if you take $(0,1]$ as a set in and of itself but viewed as a subset of $\mathbb R$, then you are working in the subset topology wherein sets like $(1/n,1]$ are open. However, if you want to view $(0,1]$ as just a subset of $\mathbb R$, then you may take the open sets $(1/n, 1+\epsilon)$ for any positive $\epsilon$.
A: As others pointed out, $(0,\frac{n-1}{n})$ is not a cover of $(0,1]$. Consider how 1 will be covered this way. You may try the case $(0,1)$ first, and see why finite many covers is not enough in this case. Then $(0,1]$ follows by the same logic. 
A: Another approach: a (subset of) a metric space* is compact if every sequence has a convergent subsequence. Consider the sequence $\{1/n:n=1,2,\ldots\}$. Does this equence have a convergent subsequence?
*With the subspace metric. This is then itself a metric space.
