I accidentally unset $PATH. Why did `echo` still work?

echo is a bash builtin. It does not use the $PATH to find the echo program, instead bash has it's own version of echo which is used instead of the echo program located in your $PATH

read more here: Bash Builtins (Bash Reference Manual)

In addition to what Minijack mentioned, you can check what a command is by using the type builtin.

$ type echo
echo is a shell builtin

On the other hand, which can be used to check for executables specifically. Once you unset $PATH, you'll get something like

$ which echo
/usr/bin/which: no echo in ((null))

Whereas with your path set you get

/usr/bin/echo

You can check man builtins for list and desription of various builtins. For example, [ and test are also builtins.

EDIT: which works for me even without PATH because of an alias that uses an absolute path

The model employed by the Single Unix Specification (a.k.a. IEEE 1003.1) is that whether a command is built into a shell is a mere optimization, and the behaviour (if invoked in a conformant manner) should be the same for a built-in version of a command as for an external version. (This is regular built-ins, that is. Special built-ins are another matter.) In particular, if the command is not found as an external in a PATH search, the built-in version is not to be executed.

This is indeed what happens with one shell. You'll find that the Watanabe shell conforms to the SUS. In its posixly-correct mode, if echo is not on the path, the built-in echo command in the Watanabe shell will not be executed.

But the 93 Korn, Debian Almquist, Z, and Bourne Again shells in their most conformant modes still all execute built-ins even if there is no corresponding executable on PATH. That is what is happening here for you. The Bourne Again shell has a built-in echo command, and several others besides. It is executing that, even though it has not found an external echo command in a PATH search.

(Note that there are quite a few ways to invoke echo in a non-conformant manner, or at least in a manner where the result is unspecified: Using -n; using a backslash in any argument; using -e; using other things expecting them to be command options or end of options markers. These not only reveal whether it is a built-in echo, but even to an extent reveal what shell is in use. Fortunately, you did not hit any of them. ☺)

Further reading

• https://unix.stackexchange.com/a/496377/5132
• https://unix.stackexchange.com/a/496291/5132
• "Command search and execution". Shell Command Language. IEEE Std 1003.1-2017. The Open Group.
• echo. Utilities. IEEE Std 1003.1-2017. The Open Group.
• "Why is printf better than echo?"