keeping a session open with urllib in python 3.6 - python-3.x

Question

I'm trying to login scrape tumblr but as when you log into the website normally through a browser, it kind of has two steps (enter email first and it check whether there is an account associated with that email then you can enter your password if the email is correct). Unfortunately, this flags up some problems when trying to automate this login without using the requests module (I'm trying to do it using urllib.request and urllib.parse which are already available in python 3.6) as there is no explicit way to start a session so you can keep the same session for the email verification and then entering the email.
Do I need to use cookies to do this or will I have to install the requests module? My code so far looks a bit like this:
import urllib.request
import urllib.parse
from html.parser import HTMLParser
input_tags = []
class myHTMLParser(HTMLParser):
def handle_starttag(self, tag, attrs):
if tag == "input":
for i in range(len(attrs)):
if attrs[i][0] == "name" and attrs[i][1] == "form_key":
input_tags.append(attrs[i+1][1])
parser = myHTMLParser()
form_key = ""
def get_form_key():
global form_key
global input_tags
url = "https://www.tumblr.com/login"
req = urllib.request.Request(url)
resp = urllib.request.urlopen(req)
resp = resp.read()
parser.feed(str(resp))
print(input_tags)
form_key = input_tags
print("form key is : ", form_key)
if len(form_key) > 1:
form_key = form_key[:1]
print("\nform key should be one value long now: ", form_key)
get_form_key()
headers = {}
headers["User-Agent"] = "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/63.0.3239.108 Safari/537.36"
url = "https://www.tumblr.com/login"
login_data = {
"determine_email" : "my.email#email.com",
"user[email]" : "my.email#email.com",
"user[password]" : "secretpassword",
"tumblrlog[name]" : "",
"user[age]" : "",
"http_referer" : "https://www.tumblr.com/logout",
"form_key" : form_key
}
encoded_data = urllib.parse.urlencode(data)
encoded_data = encoded_data.encode("utf-8")
request = urllib.request.Request(url, headers = headers, data = encoded_data)
response = urllib.request.urlopen(request)
response_url = response.geturl()
print(response_url)
This prints out the form key twice (not that important that's just from me bug checking) and then it returns the url:
https://www.tumblr.com/login
which indicates that the loin was not successful.
Any idea how to fix this?