Is the smallest non-1 divisor of a number always prime?

Let $n\in\Bbb N$ and let $a\in\Bbb N\setminus\{1\}$ be the smallest divisor of $n$. If $a$ is prime, you're done. If not, then $a$ is composite, and so can be written $a = bc$ for $b,c\in\Bbb N\setminus\{1\}$. But then $b,c < a$, and $b\mid n$, contradicting the assumption that $a$ was the smallest divisor.

No contradiction: just use the definition of prime number.

Consider the set $D=\{a\in\mathbb{N}:a\mid n, a>1\}$. The set is non empty, because $n\in D$ (assuming $n>1$, of course).

Let $p=\min D$. If $x\mid p$, then $x\mid n$ and $x\le p$. If also $x>1$, then $x\in D$. By minimality of $p$, we conclude $x=p$. Therefore $p$ is prime.

One can also avoid a proof by contradiction. By the fundamental theorem of arithmetic, every divisor $d > 1$ of $n$ is a product of some primes dividing $n$.

Let $p$ the smallest prime divisor of $n$. By the above, all divisors $d > 1$ of $n$ are greater than or equal to $p$, as every such $d$ is a product of factors all greater than or equal to $p$.

If you just want to prove that the smallest non 1 divisor of a number is always prime, all you have to do is proceed by contradiction.

Let N be a composite number ( even or odd doesn't matter ). Let's assume that N's smallest non 1 divisor is not a prime, let's call this composite divisor c. If c is composite then using the fundamental theorem of arithmetic, c can be expressed as the factorisation of prime numbers all < to c. Let p be one of c's factors. Then since c | N and p | c then p | N , but p < c ( since p is a factor of c ), which contradicts the assumption that the smallest divisor of N is c.

I don't know if this is what you wanted and if I have been clear enough.