Blinking twelve

HTML/CSS, 131 108 106 101 + 18 17 = 149 126 125 123 118 bytes

a{background:#FFF;margin:-5ch;animation:a 1s steps(2,start)infinite}@keyframes a{to{visibility:hidden

<tt>--:--<a>12:00

Edit: Saved 23 bytes thanks to @insertusernamehere. Saved 1 byte by switching from <pre> to <tt>. Saved 2 bytes thanks to @darrylyeo. Saved 5 bytes thanks to @DBS.

Octave, 63 62 61 55 bytes

c='--:--00:21';while c=flip(c)disp(c(1:5));pause(.5)end

Saved two bytes thanks to Tom Carpenter! Using a single string instead of two strings in a cell array was shorter.

Explanation:

c='--:--00:21';  % A string where the 5 first characters are --:-- or 12:00 when reversed
while c=flip(c)  % A string is always considered true in Octave, so this will loop forever
                 % while flipping the order of the string each time
 disp(c(1:5)     % display the 5 first characters of the string c
 pause(.5)       % Pause 0.5 second
end              % End loop

A few bytes saved because Octave doesn't require a colon or semicolon between flip(c) and disp(), and between pause(.5) and end.

Python2, 80 75 73 69 67 66 bytes

import time
n=0
while[time.sleep(.5)]:print"1-2-::0-0-"[n::2];n^=1

I noticed that my string magic got a little longer than picking the string from an array. Nevermind, figured it out.

Explanation:

• I set a counter n to 0, which will be toggled between 0 and 1.

• I loop endlessly with the loop while 1.

• I create a string 1-2-::0-0-, which contains the string 12:00 and --:-- interlapped.

  • Starting from the index 0 with a step of 2, we get: 12:00

  • Starting from the index 1 with a step of 2, we get: --:--

• I use n to make the repeated sequence 0, 1, 0, 1, 0... which will be the starting index of the string.

  • Using n^=1, in each loop, we get that sequence. ^ being the XOR-operator.
  • If n == 0 --> n^=1 results in 1
  • If n == 1 --> n^=1 results in 0
• I print the string, and sleep (.5 -> 0.5) seconds.

@FlipTack saved 4 bytes! --> Put the loop in one line.

@Rod saved 2 bytes! --> n+=1 to n^=1, thus n%2 to n.

@xnor saved a byte! --> while 1 --> while[time.sleep(.5)].