Aaron Swartz's xmltramp is good software. It's the simplest way I know to handle XML content. Sort of like DOM but without all the obnoxious function calls.

rssFeed = urllib.urlopen("http://.../index.rss")
rss = xmltramp.parse(rssFeed.read())

print rss.channel.title
for i in rss.channel:
  if i._name == 'item':
    print i.title

techgood
  2003-06-14 18:26 Z