I think that the answer to your question ought to be "not always".
There are examples of non-shellable balls and spheres. A very readable account of these, together with some of the history, can be found at this blog entry.
Now if you take the face lattice of a sphere, that is an Eulerian lattice. Passing from a complex $\Delta$ to the order complex of the face lattice of $\Delta$ yields the barycentric subdivision of $\Delta$.
So to find a negative answer to your question, find a non-shellable sphere whose barycentric subdivision also fails to be shellable. The construction of such spheres is briefly discussed in this paper. There is likely a source somewhere that constructs such a sphere more explicitly.
Update: Also, the answer to your pre-question is "yes". Indeed, the barycentric subdivision of any shellable complex is vertex-decomposable, a somewhat stronger property.