Partitions of a list

GolfScript, 51 characters

{[[]]\{[.;]`{1$[1$]+@@`1$`{[2$]-@@[+]+}++/}+%}/}:P;

The script defines a variable P which takes an array from top of the stack and pushes back a list of all partitions, e.g.

[1 2] P            # => [[[1] [2]] [[1 2]]]
["a" "b" "c"] P    # => [[["a"] ["b"] ["c"]] [["b"] ["a" "c"]] [["a"] ["b" "c"]] [["a" "b"] ["c"]] [["a" "b" "c"]]]

It does also work on larger lists:

6, P ,p            # prints 203, i.e. Bell number B6
8, P ,p            # 4140

You may perform own tests online.

J, 51 characters

([:<a:-.~])"1~.((>:@i.#:i.@!)#l)<@;/."1[l=:;:1!:1[1

Takes input from the keyboard, items separated by spaces:

   ([:<a:-.~])"1~.((>:@i.#:i.@!)#l)<@;/."1[l=:;:1!:1[1
a b c
+-----+------+------+------+-------+
|+---+|+--+-+|+--+-+|+-+--+|+-+-+-+|
||abc|||ab|c|||ac|b|||a|bc|||a|b|c||
|+---+|+--+-+|+--+-+|+-+--+|+-+-+-+|
+-----+------+------+------+-------+

GolfScript (43 chars)

{[[]]:E\{:v;{:^E+1/{^1$-\[~[v]+]+}/}%}/}:P;

or

{[[]]:E\{:v;{:^E+1/{^1$-\{[v]+}%+}/}%}/}:P;

Same input format, output format, and function name as Howard's solution. There's no brute forcing: this takes the simple iterative approach of adding one element from the input list to the partition each time round the outer loop.