There are a lot of approaches that can be used to establish this, but I'll try to stick to pushing your own idea through. I'm going to introduce a bit of notation first:
If we let $\varphi(n)$ be the number of positive integers less than or equal to $n$ that are relatively prime to $n$, you are saying that since for any given $n$ primes must fall among the $\varphi(n)$ congruence classes relatively prime to $n$ (except perhaps for the finite number of primes that divide $n$, and those don't affect the eventual distribution), that shows that the density of primes is at most $\frac{\varphi(n)}{n}$ for every $n$. That is,
$$\lim_{x\to\infty}\frac{\pi(x)}{x} \leq \frac{\varphi(n)}{n}\quad\text{for all }n;$$
so it suffices to show that
$$\inf\left\{\left.\frac{\varphi(n)}{n}\right|\; n\geq 0\right\} = 0.$$
For this, it is enough to show that $\liminf\frac{\varphi(n)}{n} = 0$.
This approach can work, though, and it can be done precisely by focusing on the integers $n$ that are "the product of all primes up to some $N$", so you are definitely on the right track and very close.
(I don't know if you can construct a sequence of numbers $N_k$ such that $\varphi(N_k)$ is constant and $N_k\to\infty$ as $k\to\infty$, which is what you say you want to do in your paragraph that begins "But I'm not sure what I can conclude from this..."; frankly, I doubt it can be done easily. Or at least, I can't think of a way to do it. Added: In fact, for every $K$, there is at most finitely many $n$ for which $\varphi(n)\leq K$; see below the break).
One way to push it through all the way it is to consider the following formula for $\varphi(n)$:
$$\varphi(n) = n\prod_{\stackrel{p|n}{p\text{ prime}}} \left(1 - \frac{1}{p}\right).$$
Verify that this is true.
This means that
$$\frac{\varphi(n)}{n} = \prod_{\stackrel{p|n}{p\text{ prime}}}\left(1 - \frac{1}{p}\right).$$
Now, pick $n$ to be "the product of all primes up to $N$", for some $N$. We're going to show that for the sequence we get from these very special $n$, we have $\lim\limits_{n\to\infty}\frac{\varphi(n)}{n} = 0$, which will show what we want to show.
For such an $n$ we have
$$\frac{\varphi(n)}{n} = \prod_{\stackrel{p|n}{p\text{ prime}}}\left(1 - \frac{1}{p}\right) = \prod_{\stackrel{p\leq N}{p\text{ prime}}}\left(1 - \frac{1}{p}\right).$$
So it will suffice to show that
$$\lim_{N\to\infty} \prod_{\stackrel{p\leq N}{p\text{ prime}}}\left(1 - \frac{1}{p}\right) = 0.$$
There are two pieces of information, both from Calculus, that you can use to establish this. First: if $|r|\lt 1$, then
$$\frac{1}{1-r} = 1 + r + r^2 + r^3 + \cdots + r^n + \cdots$$
and in particular, if $p$ is prime, then
$$\frac{1}{1 - \frac{1}{p}} = 1 + \frac{1}{p} + \frac{1}{p^2} + \cdots + \frac{1}{p^n} + \cdots.$$
The second piece of information is an upper bound for the integral $\int_1^k\frac{du}{u}$. Since $y = \frac{1}{u}$ is strictly decreasing, dividing the interval $[1,k]$ into $k-1$ equal parts and taking a left hand sum estimate gives that
$$\int_1^k\frac{du}{u} \lt 1 + \frac{1}{2} + \cdots + \frac{1}{k-1}.$$
Finally, one last trick: instead of looking at
$$\prod_{\stackrel{p\leq N}{p\text{ prime}}}\left(1 - \frac{1}{p}\right),$$
look at its reciprocal:
$$\frac{1}{\prod\limits_{\stackrel{p\leq N}{p\text{ prime}}}\left(1 - \frac{1}{p}\right)} = \prod_{\stackrel{p\leq N}{p\text{ prime}}}\frac{1}{1 - \frac{1}{p}} = \prod_{\stackrel{p\leq N}{p\text{ prime}}}\left(1 + \frac{1}{p} + \frac{1}{p^2} + \cdots + \frac{1}{p^k} + \cdots\right).$$
See if you can show that this is greater than or equal to a quantity which you know goes to $\infty$ as $N\to\infty$ (say, by considering the integral I mentioned above). Then you can leverage that to show the limit inferior of $\frac{\varphi(n)}{n}$ is indeed equal to $0$.
Added. In fact, for every $K\gt 0$ there are at most finitely many integers $n$ such that $\varphi(n)\leq K$, so your idea of trying to find a sequence going to $\infty$ for which $\varphi(n)$ always equals $K$ cannot prosper, I'm fraid.
To see this, note that $\varphi(n)$ is multiplicative: if $\gcd(a,b)=1$, then $\varphi(ab) = \varphi(a)\varphi(b)$. Also, if $p$ is a prime, then $\varphi(p^r) = (p-1)p^{r-1}$. This completely determines the value of $\varphi$ for any $n$, if you know the prime factorization of $n$.
Now fix $K$. If $n$ is divisible by any prime $p$ with $p\gt K+1$, then $\varphi(n)\geq \varphi(p) = p-1\gt K$. If $p$ is a prime with $p\lt K+1$, then if $r$ is such that $p^{r-1}\gt K$, then $\varphi(p^r)=(p-1)p^{r-1}\geq p^{r-1}\gt K$, so any integer divisible $n$ by at least $p^r$ will have $\varphi(n)\gt K$ as well. So any integer $n$ such that $\varphi(n)\leq K$ must be divisible only by primes less than or equal to $K+1$, and for each such prime there is a largest power that can divide $n$. This means that there are only finitely many $n$ for which $\varphi(n)\leq K$.
Let $q_1,q_2,\dots,q_n$ be odd primes of the form $3k+2$. Consider the number $N$, where
$$N=3q_1q_2\dots q_n+2.$$
It is clear that none of the $q_i$ divides $N$, and that $3$ does not divide $N$.
Since $N$ is odd and greater than $1$, it is a product of one or more odd primes. We will show that at least one of these primes is of the form $3k+2$.
The prime divisors of $N$ cannot be all of the shape $3k+1$. For the product of any number of (not necessarily distinct) primes of the form $3k+1$ is itself of the form $3k+1$. But $N$ is not of the form $3k+1$. So some prime $p$ of the form $3k+2$ divides $N$. We already saw that $p$ cannot be one of $q_,\dots,q_n$. It follows that given any collection $\{q_1,\dots,q_n\}$ of primes of the form $3k+2$, there is a prime $p$ of the same form which is not in the collection. Thus the number of primes of the form $3k+2$ cannot be finite.
Best Answer
You have a good idea, don't ruin it by doing wrong statements like “$n!+1$ is prime”.
Indeed, $n!+k$ is divisible by $k$ for $2\le k\le n$, but this gives just $n-1$ consecutive composite numbers. However, there are $n$ if you consider $(n+1)!+k$, for $2\le k\le n+1$.