Is a compact union of compact spaces still compact? I'm wondering if the following statement is true : 
Let $ K $ be a compact space, $ X $ a topological space and $ \forall x $ in $K$, $L(x)$ a compact subspace of $ X $. Then $ \cup_{x \in K} L(x) $ is a compact subspace of X.
I understand that the statement is false if $K$ is not supposed to be compact and true if $K$ is finite. (two cases that are already discussed in other topics).
If it is false, can it become true when adding some hypothesis ? 
In particular I'm interested in the case where L would be continuous. But to be accurate this means to define a topology on the set of all subspaces (or compact subspaces) of X...
Does someone have a proof of this statement (eventually in a more specific case : $X$ a metric space or a normed vector space) or a counterexample ? 
 A: Your conjecture is false.  
Let K = X = [0,1].
For all x in K, let L(x) = {x} if x is rational and {0} otherwise.  
K is compact and every L(x) is a compact subset of X. 
$ \cup_{x \in K} L(x) $ is the set of rationals in [0,1] which is not compact.  
To make your conjecture true, require K to be finite.
Notice that there are countable K counter examples.  
For L(x) to be continuous, the Hausdorff metric could be useful.
A: Having a union indexed by $K$ is not going to work because the topology on $K$ plays no actual role in the statement of the conjecture (since $K$ is just an index set, then if the conjecture were true for any compact set, it would be true for any infinite set at all). 
But instead of just taking a union, you could ask the following question: given a map $Y \to K$ such that all of the fibers are compact, is $Y$ itself compact? This is a bit more like really asking for a "compact family of compact sets." Unfortunately, it is still false, since e.g. $Y$ could be a discrete uncountable space mapping bijectively to the interval.
