What is the reason for having unreserved identifiers as built-in macros in gcc?

gcc does not fully conform to any C standard by default.

Invoke it with -ansi, -std=c99, or -std=c1x and unix won't be predefined. (-std=c1x will probably become became -std=c11 in a future more recent gcc release.)

It's a bit confusing that this is documented in the separate manual for the GNU preprocessor, not in the gcc manual.

Quoting the GNU preprocessor documentation (info cpp, version 4.5):

The C standard requires that all system-specific macros be part of the "reserved namespace". All names which begin with two underscores, or an underscore and a capital letter, are reserved for the compiler and library to use as they wish. However, historically system-specific macros have had names with no special prefix; for instance, it is common to find `unix' defined on Unix systems. For all such macros, GCC provides a parallel macro with two underscores added at the beginning and the end. If `unix' is defined, `__unix__' will be defined too. There will never be more than two underscores; the parallel of `_mips' is `__mips__'.

When the `-ansi' option, or any `-std' option that requests strict conformance, is given to the compiler, all the system-specific predefined macros outside the reserved namespace are suppressed. The parallel macros, inside the reserved namespace, remain defined.

We are slowly phasing out all predefined macros which are outside the reserved namespace. You should never use them in new programs, and we encourage you to correct older code to use the parallel macros whenever you find it. We don't recommend you use the system-specific macros that are in the reserved namespace, either. It is better in the long run to check specifically for features you need, using a tool such as `autoconf'.

The current version of the manual is here.