Update3

So I'm getting so deep in the rabbit hole, totally lost focus of my objective. I started search for faster sorting, maybe write some C or Rust, but realized that already have the data I came for processed. So I'm here to show dmesg output and one final tip about the python script. The tip is: may be better to just count using dict or Counter, than sort its output using gnu sort tool. Probably sort sorts faster than python sorted buitin function.

About dmesg, it was pretty simple to find out of memory, just did a sudo dmesg | less press G to go all way down, than ? to search back, than searched for Out string. Found two of them, one for my python script and another to my sort, the one that started this question. Here is those outputs:

[1306799.058724] Out of memory: Killed process 1611241 (sort) total-vm:1131024kB, anon-rss:1049016kB, file-rss:0kB, shmem-rss:0kB, UID:1000 pgtables:2120kB oom_score_adj:0
[1306799.126218] oom_reaper: reaped process 1611241 (sort), now anon-rss:0kB, file-rss:0kB, shmem-rss:0kB
[1365682.908896] Out of memory: Killed process 1611945 (python3) total-vm:1965788kB, anon-rss:1859264kB, file-rss:0kB, shmem-rss:0kB, UID:1000 pgtables:3748kB oom_score_adj:0
[1365683.113366] oom_reaper: reaped process 1611945 (python3), now anon-rss:0kB, file-rss:0kB, shmem-rss:0kB

That's it, thank you so much for helping so far, hope it help others too.

#!/usr/bin/env python3
import sys

conter = dict()

# Create a key for each distinct line and increment according it shows up. 
for l in sys.stdin:
    conter[l] = conter.setdefault(l, 0) + 1

# Print entries sorting by tuple second item ( value ), in reverse order
for e in sorted(conter.items(), key=lambda i: i[1], reverse=True):
    k, v = e
    print(f'{v}\t{k}')

#!/usr/bin/env python3
import sys

conter = dict()

# Create a key for each distinct line and increment according it shows up. 
for l in sys.stdin:
    conter[l] = conter.setdefault(l, 0) + 1 # After Update2 note: don't do this, do just `couter[l] = conter.get(l, 0) + 1`

# Print entries sorting by tuple second item ( value ), in reverse order
for e in sorted(conter.items(), key=lambda i: i[1], reverse=True):
    k, v = e
    print(f'{v}\t{k}')