Properties of distribution function

The probability assigned to an interval is certainly not bounded by its length. For example, discrete distributions assign positive probability to intervals of length $0.$

To prove right-continuity you need countable additivity.

\begin{align} F(x) & = \Pr(X\le x) = 1 - \Pr(X>x) \\[8pt] & = 1 - \Pr(x+1 < X \text{ or } x+\tfrac 1 2 < X\le x+1 \text{ or } x+\tfrac 1 3 < X\le x + \tfrac 1 2 \text{ or } \cdots) \\[8pt] & = 1 - \big( \Pr(x+1< X) +\Pr(x+\tfrac 1 2 < X\le x+1) + \Pr(x+\tfrac 1 3< X\le x + \tfrac 1 2) + \cdots \\[8pt] & = 1 - \lim_{N\,\to\,\infty} \sum_{n\,=\,0}^N \Pr( x + \tfrac 1 {n+1} < X \le x + \tfrac 1 n) \\[8pt] & = \lim_{N\,\to\,\infty} \Pr(X\le x + \tfrac 1 {N+1}) = \lim_{N\,\to\,\infty} F(x + \tfrac 1{N+1}). \end{align}

Given $\varepsilon>0,$ find $N$ large enough so that $F(x+\tfrac 1{N+1}) < F(x)+\varepsilon, $ and then choose $\delta= 1/N.$ Then for $x < w < x+\delta,$ you have $F(x)\le F(w)< F(x)+\varepsilon.$ The point of this paragraph is that it's not just $\lim_{N\to\infty} F(x+\tfrac 1 {N+1}) = F(x),$ but $\lim_{w\,\downarrow\,x} F(w) = F(x).$

It is a basic fact that for any finite measure $\mu$ the condition $A_n$ decreasing to $A$ implies that $\mu (A_n) \to \mu (A)$. [Lebesgue measure is an infinite measure and this property fails for Lebesgue measure]. This follow from the fact that $\mu(A_n^{c}) \to \mu(A^{c})$ since $A_n^{c}$ increases to $A$ and $\mu (E^{c})=\mu (\Omega)-\mu (E)$. With this result in hand it should be easy for you to complete your arguments.

Note that $(x,x+\delta]$ decreases to empty set as $\delta$ decreases to $0$ and $(x, \infty)$ decreases to empty set as as $x$ increases to $\infty$.