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Each entry of a zip file, as read by the zipfile module, can be accessed
via a ZipInfo object. The filename attribute of ZipInfo is a string.
However, the read method of a ZipFile object expects a binary as
argument, or at least this is what I can deduct from the following behavior:
>>> import zipfile
>>> testzip = zipfile.ZipFile('test.zip')
>>> t1 = testzip.infolist()[0]
>>> t1.filename
'tést.xml'
>>> data = testzip.read(testzip.infolist()[0])
Traceback (most recent call last):
 File "<stdin>", line 1, in <module>
 File "C:\Python30\lib\zipfile.py", line 843, in read
 return self.open(name, "r", pwd).read()
 File "C:\Python30\lib\zipfile.py", line 883, in open
 % (zinfo.orig_filename, fname))
zipfile.BadZipfile: File name in directory 'tést.xml' and header
b't\x82st.xml' differ.
The test.zip file is attached as help in reproducing this error.

The error you got is caused by giving the wrong parameters. You gave a
ZipInfo object instead of a filename.
If you execute data = testzip.read(t1.filename) yo will have no problems.

If that is what is requested, then the manual entry for ZipFile.read
must be corrected, because it states:
"ZipFile.read(name[, pwd]) .... name is the name of the file in the
archive, or a ZipInfo object."
However, Eddie, you haven't tried what you suggested, because this is
what you would get:
>>> import zipfile
>>> testzip = zipfile.ZipFile('test.zip')
>>> t1 = testzip.infolist()[0]
>>> t1.filename
'tést.xml'
>>> data = testzip.read(t1.filename)
Traceback (most recent call last):
 File "<stdin>", line 1, in <module>
 File "C:\Python30\lib\zipfile.py", line 843, in read
 return self.open(name, "r", pwd).read()
 File "C:\Python30\lib\zipfile.py", line 883, in open
 % (zinfo.orig_filename, fname))
zipfile.BadZipfile: File name in directory 'tést.xml' and header
b't\x82st.xml' differ.

Sorry, my bad.
I did tried it but with the wrong version (2.5). And it worked perfectly.
So sorry again for my mistake.
Anyways, I've found the error.
The problem is caused by different encodings used when zipping.
In open, the method is comparing b't\x82st.xml' against
b't\xc3\xa9st.xml', and of course they are different.
But they are no so different, because b't\x82st.xml' is
'tést'.encode('cp437') and b't\xc3\xa9st.xml' is 'tést'.encode(utf-8).
The problem arises because the open method supposes the filename is in
utf-8 encoding, but in __init__ it realizes that the encoding depends on
the flags. 
if flags & 0x800:
 filename = filename.decode.('utf-8')
else:
 filename = filename.decode.('cp437')

In the ZIP file format, a filename is a byte string because we don't 
know the encoding. You can not guess the encoding because it's not 
stored in the ZIP file and it depends on the OS and the OS 
configuration. So t1.filename have to be a byte string and 
testzip.read() have to use bytes and not str.

Oh, I see that zipfile.py uses the following code to choose the 
filename encoding:
 if flags & 0x800:
 # UTF-8 file names extension
 filename = filename.decode('utf-8')
 else:
 # Historical ZIP filename encoding
 filename = filename.decode('cp437')
So I'm maybe wrong: the encoding is known using a flag?

Test on Ubuntu Gutsy (utf8 file system) with zip 2.32:
$ mkdir x
$ touch x/hé
$ zip -r x.zip x
 adding: x/ (stored 0%)
 adding: x/hé (stored 0%)
$ python # 3.0 trunk
>>> import zipfile
>>> testzip = zipfile.ZipFile('x.zip')
>>> testzip.infolist()[1].filename
'x/hé'
>>> print(ascii(testzip.infolist()[1].filename))
'x/h\u251c\u2310'
Using my own file parse (hachoir-wx), I can see that flags=0 and 
filename=bytes {78 2f 68 c3 a9} ("x/hé" in UTF-8).
You can try x.zip: I attached the file.

The problem is not about reading the filenames, but reading the contents
of a file with filename that has non-ascii charaters.

I read again what STINNER Victor and I think that he found another bug.
Because, when listing the filenames of that zip file, the names are not
displayed correctly. In fact
'x/h├⌐' == 'x/hé'.encode('utf-8').decode('cp437')
So, there is again a problem with encodings when reading the contents.
The problem here is that when reading one can not give the filename,
because is not a key in the NameToInfo dictionary.

Attached is a patch that solves (I hope) the initial problem, the one
from Francesco Ricciardi.

I just "discovered" that attempting to open zip member "test\file" fails where attempting to open "test/file" works. Granted the zip contains "/" not "\" characters, but using the os.path stuff (on windows) to manipulate the names before attempting to open the zip member produces "\" characters. Clearly, I could switch them back. It seems pretty clear that zipfile should do that for me, though.
A small, self-contained zip file test case is attached, being a zip that is named .py 
My testing using Python 3.1.1

I was wondering what has prevented Eddies patch from being included into python. Has nobody volunteered to verify that it works? I would be willing to do that, though I have never compiled python on any platform before.
It just seems a bit silly to me that python cannot work with zip files with unicode file names... I just now had to do 'os.system("unzip.exe ...")' because zipfile did not work for me...

This issue looks to be a duplicate of #10801 which was only fixed (33543b4e0e5d) in Python 3.2. See also #12048: similar issue in Python 3.1.

The initial problem is clearly a duplicate of issue #10801 which is now fixed in Python 3.1+ (I just backported the fix to Python 3.1).
> I just "discovered" that attempting to open zip member "test\file"
> fails where attempting to open "test/file" works. (...)
> It seems pretty clear that zipfile should do that for me, though.
@v+python: I don't think so, but others may agree with you. Please open a new issue, because it is unrelated to the initial bug report.
I'm closing this issue because the initial is now fixed.
For x.zip (UTF-8 encoded filenames with the "Unicode" flag) problem, there is already the issue #10614 which handles this case.