Extending continuous function via Kan extension. A very classical problem in topology is the following:
Consider a topological space space $X$ and a subspace $ A \hookrightarrow X$. Suppose you have a continuous function $A \to Y$, can we extend to a function $X \to Y$?
Now we identify the topological space with the category if its open sets, thus we get:
Suppose you have functors $X \to A$ and $Y \to A$, can we find a functor $Y \to X$ closing the diagram?


*

*Is it possible to solve this problem using Kan Extension?

*Topologically speaking what does it mean for X to be cocomplete? 

*What does it mean for $X \to A$ to be full and faithful?

 A: The poset of open sets of any space is cocomplete, since open sets are closed under union, and in fact complete, since a poset is complete if and only if it's cocomplete-the meet is given by the interior of the set-theoretic intersection.
Your diagram asks for a case of Kan lifting, not Kan extension. It seems to me that the left and right Kan liftings in this case send an open subset $U$ of $Y$ to the largest open subset of $X$ whose intersection with $A$ maps into $Y$, respectively, to the smallest open subset containing the inverse image of $U$.
EDIT: This does give an actual lift of $Y\to A$ on the frame level, since for every open $V$ in $A$ there's an open $U$ in $X$ with $U\cap A=V$. But our map $Y\to X$ does not generally come from a continuous function-as it cannot, or every continuous function would extend. For sober spaces, the maps of posets corresponding to maps of spaces are those preserving finite meets and infinitary joins. In fact it seems that the lift $Y\to X$ I constructed does preserve all finite meets and all joins...of at least two elements! But it doesn't preserve the initial object. For instance, if $A\to X$ is the inclusion of the circle in the disk and $Y=A$, then the lift sends the empty set to the largest open subset of the disk whose intersection with the circle is empty, that is, to the interior of the disk, so doesn't correspond to a continuous map.
$X\to A$ being fully faithful means that when $U\cap A=V\cap A$, we must have $U=V$. (That's fullness, while faithfulness is automatic for any functor of posets.) At least for spaces with closed points this implies $X=A$, as if $a\notin A$ then $U\setminus \{a\}\cap A=U\cap A$.
A: Not an answer, but a too-long comment.
I bet that buried in the literature of categorical-topology-people there are a few elements to address an answer. 
As Kevin points out clearly you're requesting a Kan lift, and the problem is that there is no such a thing as a "pointwise Kan lift"; but this thread and the counterexample are kinda instructive, since they tell you that things get a little bit better in the proarrow equipment of profunctors ${\cal V}{\bf Cat} \to{\cal V}{\bf Prof}$, where you have a formula to compute the right Kan lift of a profunctor along another, i.e.
$$
\text{Rift}_QP \cong \int_x [{\bf X}, {\cal V}](Q(-,x), P(-,x))
$$
($\cal V$-natural transformations from $Q(-,x)$ to $P(-,x)$.)
You shall think to the 2-functor $(-)_* : {\cal V}{\bf Cat} \to{\cal V}{\bf Prof}$ as a kind of "embedding" of categories into a 2-category (semi-)freely giving an adjoint to each 1-cell:


*

*$(-)_*$ is locally fully faithful

*every $\cal V$-functor $\phi$ can be regarded as a 1-cell $\phi_*$ in ${\cal V}{\bf Prof}$ and there it has a right adjoint 1-cell.


Topological spaces are partially ordered sets, i.e. objects of the 2-category $\{0 < 1\}{\bf Cat}$. Some people like similar stuff (although in less generality and with a different purpose, I expect some good fruits to come from a careful reading of this and related sources).
