I have read the following theorem:
Let $ \{ A_{j} \} $ be a family of subsets of topological space $X$ which is either an open covering or a finite closed covering of $X$. Let $H$ be a subset of $X$ and let $ H_j := A_{j} \cap H $, for each index $j$. Then $H$ is open in $X$ if $ H_j $ is open in $A_j$ for each $j$, while $H$ is closed in $X$ if $H_j$ is closed in $A_j$ for each $j$.
The following proof is provided,
Proof: Suppose that $ \{ A_j \} $ is an open covering. If $H_j$ is open in $A_j$ for each $j$ then $H_j$ is open in $X$ for each $j$ and so the union $H$ of $H_j$ is open in $X$...
I would like to stop here. My question is that, since for each $j$, the open sets in $A_j$ should be defined as exactly those sets $ A_j \cap U $ where $U$ is open in $X$. Why can't we claim immediately that $H$ is open in $ X $ if $ A_{j} \cap H $ is open in $A_j$. Moreover, how can it be claimed that if $H_j$ is open in $A_j$ for each $j$ then $H_j$ is open in $X$ for each $j$?
Thanks in advance.
If $O$ is open in $X$ and $A$ is open in $O$, $A$ is open in $X$. (open in open is open); the proof is immediate: $A$ is open in $O$ means $A = U \cap O$ for some $U$ which is open in $X$, but then $A$ is open in $X$ as the intersection of two open sets of $X$ (namely $U$ and $O$).
If you just know that $H \cap A_j$ is open in $A_j$ for some $j$ then $H$ could be non-open: e.g. $H = [1,3]$, $A_j = (0,1)$, then $H \cap A_j = (0,1)$ is open in $(0,1)$ but $H$ is not open. We need that the $A_j$ form a cover of $X$ and that $H \cap A_j$ is open for all $j$, to conclude that $H$ is open. But the proof starts by noting that all sets $H \cap A_j$ are open in $X$, and $H$ is their union.
The same holds for closed in closed too, with the same proof, essentially.