Show that $\mathcal{S}$ is a subbasis on $\mathbb{N}$ I was given a problem:
(Edited) 

Show $$ S=\bigl\{S_p:p\in\mathbb{P}\bigr\}\cup\bigl\{\{1\}\bigr\} $$
  where $\mathbb{P}$ is the set of prime numbers, and $S = \{n \in
 \mathbb{N}: n \text{ is a multiple of }  p\}$ is a subbasis on
  $\mathbb{N}$

The easiest way to do this in my opinion to show that the $\mathcal{S}$ covers $\mathbb{N}$
i.e. $\bigcup S= \mathbb{N}$
Does this actually cover $\mathbb{N}$?
Take $n' \in \mathbb{N}$, then we need to show that $n' \in\bigcup S$
Is there some result that all natural number are multiple of primes? Without knowing that I don't see how I can show that $\mathcal{S}$ covers $\mathbb{N}$
 A: It's actually really easy to prove every number has a prime divisor, and the proof is rather nice:

Take a number $n\in\mathbb N$. If $n$ is prime, then $n=1\cdot n$ and $n$ is a multiple of a prime. 
If $n$ is not prime, then $n$ has a divisor that is not $1$ and is not $n$ (because otherwise it would be prime). Call that divisor $n_1$.
If $n_1$ is prime, then $n$ is a multiple of $n_1$, therefore a multiple of a prime. If not, $n_1$ has a divisor that is not $1$ and not $n_1$. Call that divisor $n_2$.
Continue.

Now, the process above either stops, or it does not. If it stops, it stops after it finds a divisor for $n$, so that's OK.
Now, if it doesn't stop, it produces an infinite sequence of numbers $n>n_1>n_2>n_3\dots > 1$, which is of course impossible (there can be only $n-1$ distinct numbers between $n$ and $1$!)
Conclusion: the process above stops. Therefore, $n$ has a prime divisor, or in other words, $n$ is a multiple of a prime.
A: The formulation is quite ambiguous; I believe your set $S$ consists of the sets $\{1\}$ and of the sets $S_p=\{n\in\mathbb{N}: p\mid n\}$ as $p$ varies through the prime numbers. In other terms,
$$
S=\bigl\{S_p:p\in\mathbb{P}\bigr\}\cup\bigl\{\{1\}\bigr\}
$$
where $\mathbb{P}$ is the set of prime numbers.
You want to show that
$$
\mathbb{N}=\bigcup S=\{1\}\cup\bigcup_{p\in\mathbb{P}}S_p
$$
Suppose $n\in\mathbb{N}$. If $n=1$ then $n\in\bigcup S$. Otherwise $n$ is divisible by a prime $p$, so $n\in S_p$ and $n\in\bigcup S$ as well.

How do you show that $n\ne1$ is divisible by at least a prime? If $n=0$, it's obvious. Otherwise $n>1$; consider the set
$$
D'(n)=\{k\in\mathbb{N}:k>1,k\mid n\}
$$
The set is not empty because $n\mid n$ and $n>1$. Let $p$ be the smallest element in $D'(n)$; then $p$ is prime.
