You should pass the file contents (i.e. a string) to json.loads(), not the file object itself. Try this:

with open(file_path) as f:
    data = json.loads(f.read())
    print(data[0]['text'])

There's also the json.load() function which accepts a file object and does the f.read() part for you under the hood.

Use:

json.dump(standardcategories, f, indent=2)

Instead of:

json.dump(f, standardcategories, indent=2)

Notice that I've interchanged the order of f and standardcategories

You're trying to call the file handle as a dictionary instead of the JSON dictionary stored in CONFIG. Instead, try:

with open("config.json", "r") as infile:
    try:
        CONFIG = json.load(infile)
        token = CONFIG["token"]
        owner = CONFIG["owner"]

    except (KeyError, FileNotFoundError):
        raise EnvironmentError(
            "Your config.json file is either missing, or incomplete. Check your config.json and ensure it has the keys 'token' and 'owner_id'"
        )

Note also that underscores are usually used as variable names if they won't be used anywhere, and that the underscore would be assigned to CONFIG['token'], then immediately reassigned to CONFIG['owner'] in your case. I gave them some new unique variable names if you're planning to use them later.

You shouldn't call you're variable as json. This caused the mistake.

Do something like:

import json

not_json = open('file.json')
data = json.load(not_json)

Or even better:

with open('file.json') as input_file:
    data = json.load(input_file)

[edit]

To address to what commented below, when you declare something like:

import json

json = 2

the word json now points to a number , and you lose the name of the imported package. This is true unless you declare it in a different scope (inside a function for example).

Best practice: Try to avoid naming vars / function / classes etc in names which are built-ins, known packages etc (like: list = 2; list([1,2,3]))

As per the json you have shared above, the element in the list is a string. So you need to convert that into a python dict first. You can make use of python's json module for that in the following manner:

import json
feeds = []
i = 0
for feed in output['posts']:
    feed = json.loads(feed)
    feed['id'] = i
    print(feed['id'], str(feed['title']))
    i += 1
    feeds.append(feed)

json.load() is for loading a file. json.loads() works with strings.