Reference or counter-example for Closed Graph Theorem for multivalued maps in general topological spaces Could someone be so kind to point me in the direction of a citeable proof of the following version of the Closed Graph Theorem? (i.e. assuming this is true, could someone give me a literature reference, or if it is false, a counter-example.))
Suppose $A$ and $B$ are topological spaces. For the correspondence $f : A \to B$ with closed values (i.e. $f(a)$ - closed for all $a \in A$ and $f(a)\neq\emptyset$), closed domain and compact range, to be upper hemicontinuous it is sufficient and necessary for $f$ to have closed graph. 
Edited to add: $B$ is not assumed Hausdorff.
 A: Since this result is mentioned in Wikipedia article on hemicontinuity, the references mentioned there might be a reasonable place to look for this result.
If you simply try searching for "upper hemicontinuous" "closed graph" in Google Books, there are several reasonably looking results. (Of course, you can also try Google and Google Scholar.) Notice that some authors use the name *upper semicontinuous instead of upper hemicontinuous. This gives us some other reasonable search phrases - again you can try Google Books, Google or Google Scholar.

Aliprantis C.D., Border K.C. Infinite-dimensional analysis (3ed., Springer, 2005), page 561:

17.9 Definition A correspondence $\varphi \colon X \twoheadrightarrow Y$ between topological spaces is
  closed, or has closed graph, if its graph 
  $$\operatorname{Gr}\varphi = \{(x,y)\in X\times Y; y\in\varphi(x)\}$$
  is a closed subset of $X\times Y$. 
17.11 Closed Graph Theorem A correspondence with compact Hausdorff range space is closed if and only if it is upper hemicontinuous and closed-valued. 


If $\varphi$ is single-valued (i.e., $|\varphi(x)|=1$ for each $x$), then the definition of upper hemicontinuity becomes continuity and the above result says that a function with compact Hausdorff codomain is continuous if and only if it has closed graph. (This is also sometimes called closed graph theorem.) As can be seen from counterexamples given here, the assumption that $Y$ is Hausdorff cannot be omitted in the closed graph theorem for functions. (There is an example of a function to a compact space, which is continuous and does not have closed graph.) Of course, this means that it cannot be omitted from the close graph theorem for upper hemicontinuous multifunctions, either.

EDIT: Since I made this post, the Wikipedia article linked above was edited and now (link to the current revision) it contains the following reference for the theorem in question:
J.-P. Aubin, H. Frankowska: Set-Valued Analysis (Birkhäuser, 1990), page 42

Proposition 1.4.8. The graph of an upper semicontinuous set-valued map $F\colon X\to Y$ with closed domain and closed values is closed.
The converse is true if we assume that $Y$ is compact.

Here domain is the set $\operatorname{Dom}(F)=\{x\in X; F(x)\ne\emptyset\}$ (defined on page 34). 
