Using an input string and printing it out character by character

Use a for loop!

for (size_t i = 0; i < string_length; ++i) {
     /* print ith character of the string */
}

or, if you're still < C99¹, define size_t i; before the loop and initialize it with i = 0.

Notes:

• use scanf("%19s", String) instead to preempt buffer overflows. 19 to leave room for the terminating null byte
• in C, variabels commonly have lowercase identifiers like string or charNo. How to handle concatentation of multiple words in one identifier is up to you. Common choices are first_second and firstSecond

_{¹ as @szczurcio noted in the comments to this answer}

Approach 1

1. Get the length of the string
2. Use a loop to iterate over the length of the string.
3. In each iteration of the loop, print one character.

int len = strlen(String);
for (int i = 0; i < len; ++i )
{
   printf("%c\n", String[i]);
}

Approach 2

1. Create a pointer that points to the string.
2. Use a loop in which you increment the pointer until it points to the null character.
3. In each iteration of the loop print the character that the pointer points to.

char* cp = String;
for ( ; *cp != '\0'; ++cp )
{
   printf("%c\n", *cp);
}

    #include <stdio.h>
    #include <stdlib.h>

    int main()
    {
        char String[20];
        int CharNo = 0;
        char *c;

        printf("Enter a few characters please:\n");
        scanf("%s", String);

        c=String;
        while(*c) putchar(*c++);      

        return 0;
    }