Can an open set be covered by proper open subsets? Let $(X, \tau)$ be a topological space, let $U\in\tau$. An open cover of $U$ is a set $\{U_i \ |\ i\in I\}$ (of open sets $U_i$) whose union $\bigcup U_i$ contains $U$.

If $U\subsetneq X$, then $U$ admits an open covering by open sets $\{U_i \ |\ i\in I\}$, where an arbitrary open set $U_i$ may not be a subset of  $U$. (I think.)

Does $U$ admit an open covering by open sets $\{U_i \ |\ i\in I\}$ where every open set $U_i$ is a proper subset of $U$? Less formally: given an arbitrary open set $U$, can we find an open cover of $U$ made up only of open sets "inside" of $U$?

If the answer is no, then what are the open sets $U$ that admit such "internal" coverings?

Is the answer any different if we replace $U$ by an arbitrary closed set?
It depends. Consider $\mathbb{R}$ with the usual topology and $U = (0,1)$, then we can write $(0,1) = \bigcup_{n=2}^\infty\left(\frac1n,1\right)$, for instance. Since every open set $U\subset\mathbb{R}$ is the union of intervals, it means that every open subset of $\mathbb{R}$ can be written as the union of proper open subsets of itself. It is easy to see that the same argument can be applied to general normed vector spaces.

However, if a singleton $\{x\}$ is an open set of your space then there are no proper open subsets of $\{x\}$ to unite.
Show that $K = [0,1]$ cannot be covered by open sets that are contained by $K$.

If $U$ is open, then $U$ can be covered by $\{U\}$. If $U$ is an open singleton, then there is no cover by proper subsets.

When $U$ is open, then for all $x$ in $U$, there is an open base set $U_x$ with $x$ in $U_x$ subset $U$. Often the base sets can be chosen to be proper subsets.