We need to prove that
If $z=x+y$ is irrational, then either $x$ or $y$ is irrational.
Using contrapositive, we can instead prove
If $x$ and $y$ are rational, then $z=x+y$ is rational
Which you have proved in case (1).
The converse is not true since sum of two irrational numbers may not be irrational: for example, if $x$ and $y$ are irrational in such a way that $x=-y$, $x+y=0$ is rational.
Upper bound $\dfrac{7}{6}$
Your question describes a method of breaking the range up into sub-ranges, and then estimating the sum over a sub-range by noting that each term is $\le$ the first term. But we can do much better if we estimate this sum using a linear approximation instead. Then we only need three sub-ranges to establish the upper bound of $\frac76$.
Consider a 'graph' of the function $$f(k)=\frac{1}{k}$$ for integers $k\ge 1$. If we look at any interval $m\le k\le n$ with $1\le m\le n-2$, we can compare it with the straight-line function joining points $(m,\frac{1}{m})$ and $(n,\frac{1}{n})$. We can define this function explicitly if you want, as:
$$g(k)=\frac{1}{n-m}\left(\frac{k-m}{n}+\frac{n-k}{m}\right)$$
Then $f(m)=g(m),f(n)=g(n)$; and for $m<k<n,f(k)<g(k)$ (because $f$ is a convex function).
So $$\sum_{k=m}^nf(k)<\sum_{k=m}^ng(k)$$
But the $g(k)$ are in arithmetic progression, so the RHS is half the sum of the first and the last terms multiplied by the number of terms, i.e.
$$\frac{n-m+1}{2}\left(\frac{1}{m}+\frac{1}{n}\right)$$
We divide the range into three equal sub-ranges of size $667$. This gives us:
$$\begin{align}
\sum_{k=1001}^{3001}\frac{1}{k} & =\sum_{k=1001}^{1667}\frac{1}{k}+\sum_{k=1668}^{2334}\frac{1}{k}+\sum_{k=2335}^{3001}\frac{1}{k}\\
& <\frac{667}{2}\left(\frac{1}{1001}+\frac{1}{1667}\right)+\frac{667}{2}\left(\frac{1}{1668}+\frac{1}{2334}\right)+\frac{667}{2}\left(\frac{1}{2335}+\frac{1}{3001}\right)\\
& <\frac{667}{2}\left(\frac{1}{1000.5}+\frac{1}{1667.5}+\frac{1}{1667.5}+\frac{1}{2334.5}+\frac{1}{2334.5}+\frac{1}{3001.5}\right)\\
& =\frac13+\frac15+\frac15+\frac17+\frac17+\frac19\\
& =\frac{356}{315}\\
& <\frac{7}{6}\end{align}$$
where we have used the fact that $\frac{1}{m}+\frac{1}{n}<\frac{1}{m-\frac12}+\frac{1}{n+\frac12}$ if $m<n$, again by convexity of $f(k)$.
Note that if the last term were $\frac{1}{3000}$, we could do this with just two sub-ranges, and we would get exactly $\frac{7}{6}$ as a bound. So perhaps I have missed an easier way.
Lower bound $\dfrac{29}{27}$
This time we use the convexity of the function to estimate the sum over a sub-range from below. The sum of $\frac{1}{k}$ over the range $m\le k\le n$ is greater than the number of elements multiplied by the middle element if the range contains an odd number of elements. So we get
$$\begin{align}
\sum_{k=1001}^{3001}\frac{1}{k} & =\sum_{k=1001}^{1667}\frac{1}{k}+\sum_{k=1668}^{2334}\frac{1}{k}+\sum_{k=2335}^{3001}\frac{1}{k}\\
& > 667\left(\frac{1}{1334}+\frac{1}{2001}+\frac{1}{2668}\right)\\
& = \frac12+\frac13+\frac14\\
& = \frac{13}{12}\\
& >\frac{29}{27}
\end{align}$$
Best Answer
Since $\sqrt{n^2+1}$ is an algebraic integer for $n=1001,\dots,2000$ and algebraic integers are closed under addition, it follows that $$S=\sqrt{1001^2 + 1}+\sqrt{1002^2 + 1} \ + ... + \sqrt{2000^2 + 1}$$ is algebraic integer too. Now use the fact that any rational algebraic integer IS an integer. This contradicts the fact that $S$ is NOT an integer (as you have already shown). Hence $S$ is not rational.