You need to use from urllib.request import urlopen, also I suggest you use the with statement while opening a connection.

from urllib.request import urlopen

with urlopen("https://sunlightlabs.github.io/congress/legislators?api_key='(myapikey)") as conn:
    # dosomething

You are reading the wrong documentation or the wrong Python interpreter version. You tried to use the Python 3 library in Python 2.

Use:

import urllib2

sock = urllib2.urlopen("http://diveintopython.org/") 
htmlSource = sock.read()                            
sock.close()                                        
print htmlSource

The Python 2 urllib2 library was replaced by urllib.request in Python 3.

From the docs:

Open the URL url, which can be either a string or a Request object.

...

This function returns a file-like object with three additional methods:

• geturl() — return the URL of the resource retrieved, commonly used to determine if a redirect was followed
• info() — return the meta-information of the page, such as headers, in the form of an mimetools.Message instance (see Quick Reference to HTTP Headers)
• getcode() — return the HTTP status code of the response.

Also note that as of Python 3.0, urllib.request.urlopen() and urllib.urlopen() are equivalent.

EDIT So, to time it:

# urllib.request for < python 3.0
import urllib
import time

start = time.time()

# urllib.request.urlopen() for < python 3.0
response = urllib.urlopen('http://example.com/')
data = response.read() # a `bytes` object
end = time.time()

duration = end - start