This challenge is related to MITM (Meet-in-the-Middle Attack), and the video I found explains the idea quite clear:
https://www.youtube.com/watch?v=IZFaZDO3-GQ&t=232s
## Step-by-step writeup
1. Based on the pattern of MITM, we have to obtain a plain text and a cipher text, which can be achieve by creating a plain text up to your choice, then encode it using netcat.

┌──(kali㉿kali)-[~]
└─$  nc mercury.picoctf.net 1903
Here is the flag:
92d7e5ce84c62dfb9599691b0bb152122b224939f78d2cc83d6b03abb75d760427f2c42f62ab9be1
What data would you like to encrypt? 12345678
fa5d561ae84fd7ea

Note that even the same input (plaintext) will get different output everytime, since the keys being used are not the same.

2. Create a dictionary to store the output we get by using brute-force key.

encodes={}
decodes={}
for key in tqdm(range(999999),desc="brute forcing keys"):
    key=pad(f'{key:06}')
    DESkey=DES.new(key,DES.MODE_ECB) #Key transformation
    #key1
    enc_plain=DESkey.encrypt(plaintext)
    encodes[enc_plain]=key
    #key2
    dec_cipher=DESkey.decrypt(ciphertext)
    decodes[dec_cipher]=key

`{key:06}` means that the key will be lengthen to 6 numbers long, since the key in source code (in `generate_key()`) is 6 numbers+2 white spaces.

3. Find keys

for same_inter_text in encodes_texts.intersection(decodes_texts):
    k1=encodes[same_inter_text] #k1=key
    k2=decodes[same_inter_text]
    print(k1,k2)
    break

`.keys()` will list the keys of the dictionary. Here's an [example](https://www.w3schools.com/python/trypython.asp?filename=demo_ref_dictionary_keys).
and `set` is similar to `list`, but it won't put the same key to the set twice.

4. Now we have the key, decode the flag!

DESkey1=DES.new(k1,DES.MODE_ECB)
DESkey2=DES.new(k2,DES.MODE_ECB)
inter_flag=DESkey2.decrypt(encrypted_flag)
flag=DESkey1.decrypt(inter_flag).decode()
print(flag)

The reason why we have to `.decode()` the flag is that it's originally in binary form.
## Full solution script

from Crypto.Cipher import DES
import binascii
from tqdm import tqdm

encrypted_flag = binascii.unhexlify("92d7e5ce84c62dfb9599691b0bb152122b224939f78d2cc83d6b03abb75d760427f2c42f62ab9be1")
#since the flag will be hexlified at the end of source code

def pad(msg):
    block_len = 8
    over = len(msg) % block_len
    pad = block_len - over
    return (msg + " " * pad).encode()

plaintext=pad(binascii.unhexlify("12345678").decode())#input needs to be unhexlified, decoded, and padded (as the input in source code)
ciphertext=binascii.unhexlify("fa5d561ae84fd7ea")#same reason as "encrypted flag"

encodes={}
decodes={}
for key in tqdm(range(999999),desc="brute forcing keys"):
    key=pad(f'{key:06}')
    DESkey=DES.new(key,DES.MODE_ECB) #Key transformation(step 1)
    #key1
    enc_plain=DESkey.encrypt(plaintext)
    encodes[enc_plain]=key
    #key2
    dec_cipher=DESkey.decrypt(ciphertext)
    decodes[dec_cipher]=key

print("###finding intersection###")

encodes_texts=set(encodes.keys()) #set of "enc_plain"
decodes_texts=set(decodes.keys()) #set of "dec_cipher"



for same_inter_text in encodes_texts.intersection(decodes_texts):
    k1=encodes[same_inter_text] #k1=key
    k2=decodes[same_inter_text]
    print(k1,k2)
    break

DESkey1=DES.new(k1,DES.MODE_ECB)
DESkey2=DES.new(k2,DES.MODE_ECB)
inter_flag=DESkey2.decrypt(encrypted_flag)
flag=DESkey1.decrypt(inter_flag).decode()
print(flag)

## REFs
- https://stackoverflow.com/questions/339007/how-do-i-pad-a-string-with-zeroes
- https://www.geeksforgeeks.org/data-encryption-standard-des-set-1/
- https://www.youtube.com/watch?v=cVhlCzmb-v0
- https://www.chiragbhalodia.com/2021/09/des-algorithm.html
- https://stackoverflow.com/questions/59925060/des-encryption-in-python
- https://pycryptodome.readthedocs.io/en/latest/src/cipher/des.html
- https://www.w3schools.com/python/ref_set_intersection.asp
- https://www.w3schools.com/python/ref_dictionary_keys.asp
- https://www.w3schools.com/python/python_sets.asp
- https://wenyuangg.github.io/posts/python3/python-set.html
- https://ithelp.ithome.com.tw/articles/10249953