Does $f(x)=ax$ intersect $g(x)=\sqrt{x}$ It maybe a stupid question but I want to be sure how to explain it formally.
Does $f(x)=ax$ intersect $g(x)=\sqrt{x}$, when $x>0$ and $a>0$ (however small it is)
I think it does. 
The derivative of $f(x)$ is constant, positive. And the derivative of $g(x)$ tends to $0$. So there will be some point $x_0$, from which the derivative of $f$ will be greater than derivative of $g$. Therefore $g$ will grow slower than $f$ and both functions finally meet. Am I right? This is enough? Can one formally prove it? 

If there is a point of intersection, it will satisfy the equation $$ax = \sqrt x\implies (ax)^2 = x \iff a^2x^2 - x = 0 \iff x(a^2x - 1) = 0\;$$
Indeed, the graphs intersect, when $x = 0$ and when $a^2x-1=0 \iff x = \frac 1{a^2}$. Since we are interested in only $x\gt 0$, the point of intersection you are looking for is $$(x, f(x)) = (x, g(x))=\left(\frac 1{a^2}, g\left(\frac 1{a^2}\right)\right) = \left(\frac 1{a^2}, \frac 1a\right)$$
Note: we know that $a > 0$ and $x>0$. Hence $$g\left(\frac 1{a^2}\right) = \sqrt{\frac 1{a^2}} = \frac 1a$$ 
