Let $X$ be a set and let ${\frak T}$ be a collection of [paracompact](https://en.wikipedia.org/wiki/Paracompact_space) topologies on $X$ such that for any $\tau, \tau'\in {\frak T}$ we have $\tau\subseteq \tau'$ or $\tau'\subseteq \tau$. Let $\sigma$ be the topology having $\bigcup {\frak T}$ as a [base](https://en.wikipedia.org/wiki/Base_(topology)).


Is $(X,\sigma)$ necessarily paracompact?


**EDIT.** Thanks to Tomek Kania for observing that $\bigcup {\frak T}$ need not be a topology.



Is the topology generated by the union of a chain of paracompact topologies paracompact?
Here is an easy counterexample under the Continuum Hypothesis, where the *union of the topologies* generates a non-normal topology.
Take the upper half plane $\mathbb{H}$ (including $x$-axis) and enumerate $\mathbb{R}$ as $\langle x\_\alpha:\alpha<\omega\_1\rangle$.
Now let topology $\mathcal{T}\_\alpha$ be generated by the normal topology plus the tangent disk neighbourhoods at all points $(x\_\beta,0)$ for $\beta\le\alpha$.
Each topology $\mathcal{T}\_\alpha$ is still separable and metrizable (regular with a countable base) but the union generates the Tangent Disk topology, which is not (even) normal.


This example can be made `real' as follows: take an injective sequence $\langle x\_\alpha:\alpha<\omega\_1\rangle$ of real numbers and consider the subset $X$ of $\mathbb{H}$ consisting of all points above the $x$-axis and the points $(x\_\alpha,0)$ on the x-axis.
Perform exactly the same construction as above; the intermediate topologies are again separable and metrizable and their union generates the Tangent Disk topology on $X$. In this topology the set $\{(x\_\alpha,0):\alpha<\omega\_1\}$ is closed and discrete but there is no pairwise disjoint family $\{U\_\alpha:\alpha<\omega\_1\}$ of open sets such that $(x\_\alpha,0)\in U\_\alpha$ for all $\alpha$. So the space is not collectionwise Hausdorff and hence not paracompact.


