You build the object before encoding it to a JSON string:

import json

data = {}
data['key'] = 'value'
json_data = json.dumps(data)

JSON is a serialization format, textual data representing a structure. It is not, itself, that structure.

source : https://docs.python.org/3/library/json.html

import json
data = {"user2_proximity": 3, "Wifi_1": -80, "Wifi_2": -40, "Wifi_3": -40, 
"thermostat": 18, "light": 0, "hour_of_day": 0, "user3_proximity": 3, 
"user1_proximity": 1, "day_of_week": 1, "security": 0, "minute_of_hour": 9, 
"Act_1": 1, "Act_2": 0, "Act_3": 0}

json_data = json.dumps(data)

Just encode the end result. You are building a Python dictionary, add all your keys to one dictionary and encode the final result. Encoding to JSON then decoding again doesn't really achieve anything here.

studentInfo = {}
for i in range(1,5)
    studentInfo[id[i]] = marks[i]

jsonstring = json.dumps(studentInfo)

Rather than use an index, you can use zip() to combine id entries with marks entries:

studentInfo = {}
for student_id, student_marks in zip(id, marks):
    studentInfo[student_id] = student_marks

jsonstring = json.dumps(studentInfo)

Better still, you can simply produce the whole dictionary from the zip() output directly:

studentInfo = dict(zip(id, marks))
jsonstring = json.dumps(studentInfo)

If you're trying to create a JSON string where the value of one property is an escaped JSON string, you simply create the JSON from your first dictionary and assign it to property of your "outer" dictionary and then create the JSON from that:

import json
d = {
    "Message": json.dumps( {"fruit": "apple"} )
}
print(json.dumps(d))

Output:

{"Message": "{\"fruit\": \"apple\"}"}

Note that you can not generate your desired output of {"Message":{\"fruit\": \"apple\"}} as this would not be valid JSON.

By the way, you do not "turn JSON into a string", because by definition, JSON is already a string.