When studying "Paracompactness", I thought that
If $X$ is regular, and $X= \bigcup Y_n$ ($n\ge1$) where $Y_n$ is paracompact subspace of $X$,
then $X$ is paracompact.
Here is the proof.
Suppose $\mathcal A$ be open covering of $X$.
For each $n\ge1$, since $\mathcal A$ is open covering of $Y_n$,there exists open refinement $\mathcal B_n$ of $\mathcal A$ such that is locally finite and covers $Y_n$ .
Then define, $$\mathcal B= \bigcup \mathcal B_n$$
Obviously, $\mathcal B$ is open covering of $X$. And for any $B \in \mathcal B$,
$\exists n\ge 1$ such that $B \in \mathcal B_n$So, there exists $A \in \mathcal A$ such that $B \subset A$ , because $\mathcal B_n$ is refinement of $\mathcal A$.
Thus, $\mathcal B$ is open refinement of $\mathcal A$.Now, we have refinement $\mathcal B$ of $\mathcal A$ that is open covering of $X$ and $\sigma$-locally finite (or countably locally finite).
Because $X$ is regular, we also have locally finite open refinement of $\mathcal A$ that covers $X$. And $X$ is paracompact.
Is here any problem in proof? I think this would be wrong, but I cannot find any of it.
Best Answer
It is not clear, in your proof, why $\mathcal B_n$ would be locally-finite. Its restriction to $Y_n$ (that is, $\{B\cap Y_n:B\in\mathcal B_n\}$) is locally-finite in $Y_n$, but not necessarily in $X$.
You may take a specific example when the statement fails, and try to track exactly where your proof fails.
A standard example of a Tychonoff space that is not normal is the Moore (or Niemytzki) plane, or tangent disk space. It is the $x$-axis together with the upper half-plane. If $H$ is the $x$-axis, then the topology induced on $H$ is discrete, hence $H$ is paracompact. If $K=\{(x,y):y>0\}$ is the upper half-plane then the topology induced on $K$ is the standard one, hence $K$ is also paracompact. Yet the Moore plane $\Gamma=H\cup K$ is not normal, hence it is not paracompact.