If a function $f$ on a open set $U$ is uniformly continuous on any closed subset, then is it continuous?
I find that the uniform continunity can be “summed” on finite sets(see here). But it is a problem that $U$ may not be represented by finite cover of closed subset.
Presumably $f$ is a function from an open subset $U$ of a metric space to a metric space, otherwise "uniform continuity" doesn't mean anything. Metric spaces are regular spaces, so if $x\in U$ there is an open $V$ with $x\in V\subset \overline V\subset U.$ And $f$ restricted to $\overline V$ is continuous by hypothesis.
A function $f$ such that every $x\in dom(f)$ has a nbhd $W$ such that $f|_W$ is continuous is called a locally continuous function. (A nbhd of $x$ is any $W$ such that $x\in V\subset W$ for some open $V$.)
There are many properties of continuous functions (between any spaces) that are equivalent to continuity. One of them is:
$f$ is continuous iff $f$ is locally continuous.
Remark: If $(X,d)$ is a metric space and $x\in U\subset X$ where $U$ is open in $X$ then for some $r>0$ the open ball $B_d(x,r)$ is a subset of $U.$ Now if $V=B_d(x,r/2)$ then $x\in V\subset \overline V\subset U.$
