Analysis: Is $A$ dense in $[0,1]$? Let $f_n:[0,1]\to\mathbb R$ define by $f_n(x)=\cos(nx)$. Let $A_n=\{x\mid f_n(x)=0\}$ and $A=\bigcup_{n\in\mathbb N}A_n$. I have shown that $|A|=+\infty$. Do you think that $A$ is dense in $[0,1]$ ? I think that it is, and my proof would go like this:
Suppose by contradiction that it's not. Then there is an $\varepsilon>0$ and a $x\in [0,1]\backslash A$ such that $]x-\varepsilon,x+\varepsilon[\cap A=\emptyset.$ Then $f_n(y)\neq 0$ for all $n\in\mathbb N$ and for all $y\in]x-\varepsilon,x+\varepsilon[$. With loss of generality, suppose that $\cos(ny)>0$ for all $n\in\mathbb N$ and all $y\in]x-\varepsilon,x+\varepsilon[.$ Let $n$ enough big for that $\frac{1}{n}<\varepsilon$. Then for all $k>n$, $\cos(kx)>0$ and thus $\cos(ky)>0$ for all $y\in \left[x-\frac{1}{n},x+\frac{1}{n}\right]+\frac{2k \pi}{n}$ and so on $$[0,1]\cap \bigcup_{k\in\mathbb N}\left(\left[x-\frac{1}{n},x+\frac{1}{n}\right]+\frac{2k \pi}{n}\right)$$ which is actually a finite union what is a contradiction with the fact that $|A|=\infty $. 
Do you think that it work ?

Why approach by contradiction when you can do it directly? 
You know explicitly that $\cos(nx) = 0$ when $x = \frac{1}{n} (k+ \frac12) \pi$ for any $k\in \mathbb{Z}$. This means that any $y\in [0,1]$ is within $\pi/n$ distance of a zero of $f_n(x)$. This is enough to show that $y$ is arbitrarily close to elements of $A$. 
