Get a limit number to test all the prime numbers it contains in C

My task was:

Write a program that accepts an integer as input and then displays all the prime
numbers smaller than or equal to that number.

And my code is:

#include 
void FindPrime(int number);

int main()
{        
    int userInput, counter;
    printf("Please enter a limit number to test all the prime numbers it contains:\n");

    scanf("%d", &userInput);

    printf("The prime numbers %d contains are:\n", userInput);

    for (counter =  2; counter < userInput; counter++)        
    {
        FindPrime(counter);
    }

    return 0;
}

void FindPrime(int number)    
{
    int numberTest, nextNumber = 1;

    for (numberTest = 2; numberTest < number; numberTest++)
    {
        if (number%numberTest == 0 && numberTest < number)
            continue;
        else 
            nextNumber++;                
    }

    if (nextNumber == (number - 1))
        printf("%d ", number);
}

Can this code be improved?

-
Personally, I would put FindPrime above main, to avoid having to declare it.

-
Only have one variable declaration per line. With more than one, it becomes harder to see where a variable comes from. It doesn't really matter with a short program, but will for more complex ones.

-
Your print statements contain text which contradict what the task says (contains rather than less than)

-
userInput needs to be validated - Make sure that the user has entered a number, and it's greater than 0.

-
As mentioned, it's better that a function does one thing, rather than multiple. So your FindPrime function should be responsible for finding one, and the caller should be responsible for printing it.

-
In FindPrime, the if statement contains a condition that's already been tested for in the for loop, so you can simplify the if condition by removing it.

-
Always use {} for if (and while, for etc) statements, even if they're single line. Sooner or later, you'll add another line and break things, causing a very subtle error. And in this example, I'd reverse the condition, since the normal case doesn't do anything.

if (number%numberTest != 0 )
{
    nextNumber++;
}

With regards to the algorithm, it can certainly be improved, but that may be out of your reach at the moment.

StackExchange Code Review Q#21056, answer score: 2