You can use pointer arithmetic and the function memcpy:

#include <stdio.h>
#include <string.h>

int main( void )
{
    char str1[] = "123copy321";
    char str2[5];

    //copy str1[3] up to and including str1[6] to str2
    memcpy( str2, str1 + 3, 4 );

    //add terminating null character to str2
    str2[4] = '\0';

    printf( "%s\n", str1 );
    printf( "%s\n", str2 );
}

This program has the following output:

123copy321
copy

A C string is a nul-terminated character array.

The C language does not allow assigning the contents of an array to another
array. As noted by Barry, you must copy the individual characters one by one
from the source array to the destination array. e.g. -

#define _CRT_SECURE_NO_WARNINGS
#include 
#include 

int main()
{
    char str1[] = "Hello";
    char str2[10] = {0};

    for (int x = 0; x < strlen(str1); ++x)
    {
        str2[x] = str1[x];
    }

    printf("%s\n", str2);

    return 0;
}

To make this common task easier there are standard library functions provided
which will perform this operation. e.g. - memcpy(), etc.

memcpy(str2, str1, 6);

When the array contains a nul-terminated string of characters you can use
strcpy(), etc.

strcpy(str2, str1);

Caveat: Some of the above functions are considered unsafe as they do not guard
against buffer overruns of the source and destination arrays. There are safer
versions provided by the compiler.

Note that if and when you start learning C++ you will find that there you can
assign a C++ std::string object to another object of the same type. However,
even in C++ the same rules apply when working with C strings, "raw" character
arrays, etc.

• Wayne

You're incrementing dest during the while loop. You need to keep hold of a pointer to the start of the buffer to return from the function.

char *stringcopywithpointer( const char *source)
{
    int ii = 0;
    int len = strlen(source);
    char *copy = malloc(len+1);
    char* dest = copy;

    while(*source != '\0')
    {   
        *dest++ = *source++;
    }

    *dest = '\0';
    printf("\n copied string = %s", copy);
    return copy;
}

Note that you could save some code by using strcpy

char *stringcopywithpointer( const char *source)
{
    int len = strlen(source);
    char *copy = malloc(len+1);
    strcpy(copy, source);
    return copy;
}

and you could reduce this to a single line if you have access to the non-standard strdup

char *stringcopywithpointer( const char *source)
{
    return strdup(source);
}

First, you need add +1 in size for the \0.

char color1[3];
char color2[5];

and then:

strncpy(color1, string, 2);
color1[3] = '\0';

strncpy(color2, string + 2, 4); 
color2[4] = '\0';

Assuming that

char *string = "AAbbCC"; 

printf("color1 => %s\ncolor2 => %s\n", color1, color2);

The output is:

color1 => AA
color2 => bbCC

I hope this help you.

UPDATE

You can write a substr() function to get part of string(from x to y) and then copy to your string.

char * substr(char * s, int x, int y)
{
    char * ret = malloc(strlen(s) + 1);
    char * p = ret;
    char * q = &s[x];

    assert(ret != NULL);

    while(x  < y)
    {
        *p++ = *q++;
        x ++; 
    }

    *p++ = '\0';

    return ret;
}

Then:

char *string = "AAbbCC"; 
char color1[3];
char color2[4];
char color3[5];
char *c1 = substr(string,0,2);
char *c2 = substr(string,2,4);
char *c3 = substr(string,4,6);

strcpy(color1, c1);
strcpy(color2, c2);
strcpy(color3, c3);

printf("color1 => %s, color2 => %s, color3 => %s\n", color1, color2, color3);

The output:

color1 => AA, color2 => bb, color3 => CC

And Don't forget:

free(c1);
free(c2);
free(c3);