Is the sequence $x_n=\dfrac{1}{\sqrt{n}}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{n}}\right)$ monotone?

The sequence is indeed increasing. Using what you have left off we need to prove: $x_{n+1} - x_n > 0\iff ...x_n < 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$. We prove this by induction on $n \ge 1$. Clearly $x_1 = 1 < 1+ \sqrt{\frac{1}{2}}$. Assume $x_n < 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$, we show: $x_{n+1} < 1+\dfrac{\sqrt{n+1}}{\sqrt{n+2}}$. Using the recursive formula you had above: $x_{n+1} = \dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}< \dfrac{\sqrt{n}}{\sqrt{n+1}}\left(1+\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)+\dfrac{1}{n+1}= 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}< 1+\dfrac{\sqrt{n+1}}{\sqrt{n+2}}$ which is clear because $n(n+2) < (n+1)^2$. Thus by induction $x_n < 1 +\dfrac{\sqrt{n}}{\sqrt{n+1}}$ and in turn implies $x_{n+1} > x_n, \forall n \ge 1$. Thus the sequence is increasing.

A generalization I couldn't pass by. Let $\color{blue}{S_n=\frac1n\sum_{k=1}^{n-1}f\big(\frac{k}{n}\big)}$ where $f:(0,1)\to\mathbb{R}$ is strictly convex: $$f\big((1-t)a+tb\big)<(1-t)f(a)+tf(b)\quad\impliedby\quad a<b,0<t<1.$$ If we put $a=k/(n+1),b=(k+1)/(n+1),t=k/n$ for $0<k<n$ here, we obtain $$f\Big(\frac{k}{n}\Big)<\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\frac{k}{n}f\Big(\frac{k+1}{n+1}\Big),$$ which, after summing over $k$ (to have "$<$" still, we must assume $n>1$), gives $$\sum_{k=1}^{n-1}f\Big(\frac{k}{n}\Big)<\sum_{k=1}^{\color{red}{n}}\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\sum_{k=\color{red}{1}}^{n}\frac{k-1}{n}f\Big(\frac{k}{n+1}\Big)=\frac{n-1}{n}\sum_{k=1}^{n}f\Big(\frac{k}{n+1}\Big),$$ i.e. $\color{blue}{n^2 S_n<(n^2-1)S_{n+1}}$. Returning to the question, if we put $f(x)=1/\sqrt{x}$, we get $x_n=S_n+1/n$ and $n^2 x_n<(n^2-1)x_{n+1}+1$; since $x_{n+1}>1$, the latter implies the needed $x_n<x_{n+1}$.