Lang's proof that there exists $x>0$ such that $\cos x=0$

To the first question : if sine starts from zero and increases, it is positive.
So the derivative of cosine $g'=-f $ is negative.
Hence cosine decreases.

Added

To the second question: as cosine starts from $1$ at $x=0$ and decreases, it either reaches zero at some $x_0 > 0$, or not.
So there exists $a$ such that $1 = \cos 0 > \cos a > 0$ (that would be any $a>0$ in the latter case, or some $0<a<x_0$ in the former).
Then, the positive value less than $1$, when raised to a positive power $2^n$, results in decreasing values as $n$ grows (we get a fraction of the same fraction of the fraction, and so on).
However, it never gets negative, as a product of two positive values $(\cos a)^{2^n}\,\cdot\,(\cos a)^2$ makes a positive result.
As a result we get a strictly decreasing positive sequence of cosine's values at $x_n=2^na$.
But we know the function is strictly decreasing, so its values between $x_n<x_{n+1}$ are between $\cos(x_n)$ and $\cos(x_{n+1})$, hence positive, too, and bounded from above by decreasing $\cos(x_n) = (\cos a)^{2^n}$ as $n$ grows.
This 'squeezes' the cosine towards zero.

I wonder. however, how did author get the identity $$g(2a)=(g(a))^2−(f(a))^2$$ just form the presented assumptions...

The conditions immediately imply that the derivative of $f(x)^2+g(x)^2$ is $0$. So $f(x)^2+g(x)^2=f(0)^2+g(0)^2=1$ for all $x$.

If $f'(x)>0$ for all $x\ge 0$ then $\forall t>1\,(-f(t)<-f(1)<-f(0)=0)$. But then $$x>1\implies g(x)-g(1)=\int_1^xg'(t)dt=\int_1^x(-f(t))dt<$$ $$<\int_1^x(-f(1))dt=(x-1)(-f(1))\implies$$ $$\implies \lim_{x\to\infty}g(x)=-\infty$$ contrary to $ g(x)^2=1-f(x)^2\le 1.$

So there must exist $x_1>0$ with $f'(x_1)\le 0.$ And since $f'=g$ is continuous with $f'(0)>0$ there must exist $x_2\in (0,x_1]$ with $0=f'(x_2)=g(x_2).$