Floor plus square root of the fractional part

Inside $(n,n+1)$. Here $\lfloor x\rfloor=n$ and $\{x\}=x-n$, so $f(x)=n+\sqrt{x-n}$, which is continuous.

At an integer $n$, from the right. For $x\to n^+$ (with $x<n+1$), $f(x)=n+\sqrt{x-n}\to n=f(n)$.

At an integer $n$, from the left. For $x\to n^-$ (with $x>n-1$), $\lfloor x\rfloor=n-1$ and $\{x\}=x-(n-1)\to1$. So $f(x)\to(n-1)+\sqrt1=n=f(n)$.

Conclusion. The two one-sided limits equal $f(n)$, so $f$ is continuous at every integer, hence on all of $\mathbb{R}$.

Remark. The floor function alone jumps by $1$ at each integer, and the square root of the fractional part climbs from $0$ to $1$ just in time to compensate. In fact $f$ is also strictly increasing.