Show that the continuously differentiable sequence $N_n(x)$ is increasing, where $$N_n(x)=N(nx),\,\, \frac{d}{dx}N(x)\geq0\,\,\forall x,\,\,\,N(x)=0\,\,\forall x\leq 0, \,\, N(x)=1\,\,\forall x\geq 1$$
I tried to prove this using induction but it did not lead me anywhere. Specifically, I stated that since $N’(x)\geq 0$ then assuming it is true for $n$ it follows that $ \frac{d}{dx}N((n+1)(x))= nN’((n+1)(x))+N’((n+1)(x))$, but I am not sure how to show that this is non-negative.
I am not sure whether the given statement is effectively true because if I define the sequence as follows: $$N_n(x):=\begin{cases} (-1)^n+1,\,\,& x\leq 0\\ g(nx)\,\,&0<x<1\\ (-1)^{n+1}\,\,&x\geq 1 \end{cases}$$ and $g(nx)$ is defined to be increasing for $n$ odd and decreasing for $n$ even, then this specific definition of $N_n(x)$ shows that it is not true that $N_{n+1}(x) \geq N_n(x)$. Am I wrong?
Your method of defining $N_n$ is confused : according to the constraints, once $N(x)$ is defined, $N_n(x)$ is defined for all $n$ -- it's $N_n(x) = N(nx)$.
Note that $N(x) = 0$ for all $x \leq 0$, so $N_n(x) = N(nx) = 0$ for all $n \in \Bbb{Z}_{>0}$ and all $x \leq 0$. Consequently, since your proposed $N_n(x)$ is $(-1)^n + 1$ on $x \leq 0$, your proposed $N_n$ does not meet the given constraints. Similarly, the constraints force $N_n(x) = 1$ for $x \geq 1/n$, so your proposed $N_n$ does not meet the constraints on that piece of the function.
In fact, $N_n$ is a CDF with (closed) support (a closed subset of) $[0,1/n]$.
More to your question, $N_n'(x) = \frac{\mathrm{d}}{\mathrm{d}x} N(n x) = n N'(n x) \geq 0$ (because $nx$ is just some real number). So each $N_n(x)$ is increasing along $x$.
Furthermore, $N_{n+1}(x) - N_n(x) = N((n+1)x) - N(n x)$. For $x \leq 0$, this difference is $0$. For $x > 0$, the mean value theorem shows there is a point, $c$ between $nx$ and $(n+1)x$ where $N'(c) \geq 0$ is the same as the average rate of change of $N$ between $nx$ and $(n+1)x$, so $N_n(x)$ is also increasing along $n$.