[py] Avoid waiting indefinitely on a frozen chromedriver process (#14578)

Add a default timeout of 2 minutes two requests to the webdriver

Author: dbernhard-0x7CD

Diff for: py/selenium/webdriver/remote/remote_connection.py

@@ -305,7 +305,7 @@ def execute(self, command, params):
         LOGGER.debug("%s %s %s", command_info[0], url, str(trimmed))
         return self._request(command_info[0], url, body=data)
 
-    def _request(self, method, url, body=None):
+    def _request(self, method, url, body=None, timeout=120):
         """Send an HTTP request to the remote server.
 
         :Args:
@@ -323,12 +323,12 @@ def _request(self, method, url, body=None):
             body = None
 
         if self.keep_alive:
-            response = self._conn.request(method, url, body=body, headers=headers)
+            response = self._conn.request(method, url, body=body, headers=headers, timeout=timeout)
             statuscode = response.status
         else:
             conn = self._get_connection_manager()
             with conn as http:
-                response = http.request(method, url, body=body, headers=headers)
+                response = http.request(method, url, body=body, headers=headers, timeout=timeout)
             statuscode = response.status
         data = response.data.decode("UTF-8")
         LOGGER.debug("Remote response: status=%s | data=%s | headers=%s", response.status, data, response.headers)