Any way to exit bash script, but not quitting the terminal

Yes; you can use return instead of exit. Its main purpose is to return from a shell function, but if you use it within a source-d script, it returns from that script.

As §4.1 "Bourne Shell Builtins" of the Bash Reference Manual puts it:

     return [n]

Cause a shell function to exit with the return value n. If n is not supplied, the return value is the exit status of the last command executed in the function. This may also be used to terminate execution of a script being executed with the . (or source) builtin, returning either n or the exit status of the last command executed within the script as the exit status of the script. Any command associated with the RETURN trap is executed before execution resumes after the function or script. The return status is non-zero if return is used outside a function and not during the execution of a script by . or source.

Instead of running the script using . run2.sh, you can run it using sh run2.sh or bash run2.sh

A new sub-shell will be started, to run the script then, it will be closed at the end of the script leaving the other shell opened.

The "problem" really is that you're sourcing and not executing the script. When you source a file, its contents will be executed in the current shell, instead of spawning a subshell. So everything, including exit, will affect the current shell.

Instead of using exit, you will want to use return.

You can add an extra exit command after the return statement/command so that it works for both, executing the script from the command line and sourcing from the terminal.

Example exit code in the script:

   if [ $# -lt 2 ]; then
     echo "Needs at least two arguments"
     return 1 2>/dev/null
     exit 1
   fi

The line with the exit command will not be called when you source the script after the return command.

When you execute the script, return command gives an error. So, we suppress the error message by forwarding it to /dev/null.