$\newcommand\ep\epsilon\newcommand\Om\Omega\newcommand\al\alpha\newcommand\R{\mathbb R}$According to your comment, $C^{0,2s+\ep}(\Om)$ is the set of all functions on $\Om$ that are Hölder-continuous with exponent $2s+\ep\in(0,1)$. It also appears that $\ep>0$ and that you extend the functions $u\in C^{0,2s+\ep}(\Omega)$ to $\R^n$ by letting $u:=0$ on $\R^n\setminus\Om$.

Take any such extended function $u$ that is Hölder-continuous on $\R^n$ with exponent $2s+\ep$. Taking into account the boundedness of $\Om$, we see that $$|u(x)-u(y)|\le\min(c,c|x-y|^{2s+\ep})$$ for some real $c>0$ and all $x,y$ in $\R^n$. Then for all $x\in\Om$, letting $B_x$ denote the (say open) unit ball in $\R^n$ centered at $x$, we have $$\int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy=I_1+I_2,$$ where $$I_1:=\int_{B_x}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy \le c\int_{B_x}\frac{|x-y|^{2s+\ep}}{|x-y|^{n+2s}}\,dy \asymp\int_0^1 \frac{r^{2s+\ep}}{r^{n+2s}}r^{n-1}\,dr=\int_0^1 r^{\ep-1}\,dr<\infty$$ and $$I_2:=\int_{\R^n\setminus B_x}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy \le c\int_{\R^n\setminus B_x}\frac{dy}{|x-y|^{n+2s}} \asymp\int_1^\infty\frac{r^{n-1}\,dr}{r^{n+2s}} =\int_1^\infty r^{-1-2s}\,dr<\infty.$$ So, $$\sup_{x\in\Om}\int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy<\infty,$$ as desired.

On the other hand, if the extended function $u$ is not assumed to be Hölder-continuous on the entire space $\R^n$, then your desired conclusion will not hold in general. E.g., let $\Om$ be the open unit $\ell^\infty$-ball in $\R^n$ and let $$u(x):=\|x\|_\infty^{2s+\ep}\,1(x\in\Om)=\|x\|_\infty^{2s+\ep}\,1(\|x\|_\infty<1)$$ for all $x\in\R^n$. Then $u$ is Hölder-continuous on $\Om$ with exponent $2s+\ep$. However, for $x=(a,0,\dots,0)\in\Om$ with $a\uparrow1$ $$ \begin{aligned} \int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy &\ge\int_1^\infty dy_1\int_{\R^{n-1}}\, \frac{a^{2s+\ep}\,dy_2\cdots dy_n}{((y_1-a)^2+\sum_{j=2}^n y_j^2)^{n/2+s}} \\ &\to\int_1^\infty dy_1\int_{\R^{n-1}}\, \frac{dy_2\cdots dy_n}{((y_1-1)^2+\sum_{j=2}^n y_j^2)^{n/2+s}} \\ &=\frac12\,\int_{\R^n}\frac{dz}{|z|^{n+2s}} \\ &\asymp\int_0^\infty \frac{r^{n-1}\,dr}{r^{n+2s}}=\int_0^\infty r^{-1-2s}\,dr=\infty. \end{aligned} $$ So, here $\int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy$ is unbounded.

$\newcommand\ep\epsilon\newcommand\Om\Omega\newcommand\al\alpha\newcommand\R{\mathbb R}$According to your comment, $C^{0,2s+\ep}(\Om)$ is the set of all functions on $\Om$ that are Hölder-continuous with exponent $2s+\ep\in(0,1)$. It also appears that $\ep>0$ and that you extend the functions $u\in C^{0,2s+\ep}(\Omega)$ to $\R^n$ by letting $u:=0$ on $\R^n\setminus\Om$.

Take any such extended function $u$ that is Hölder-continuous on $\R^n$ with exponent $2s+\ep$. Taking into account the boundedness of $\Om$, we see that $$|u(x)-u(y)|\le\min(c,c|x-y|^{2s+\ep})$$ for some real $c>0$ and all $x,y$ in $\R^n$. Then for all $x\in\Om$, letting $B_x$ denote the (say open) unit ball in $\R^n$ centered at $x$, we have $$\int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy=I_1+I_2,$$ where $$\begin{aligned}I_1&:=\int_{B_x}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy \\ &\le c\int_{B_x}\frac{|x-y|^{2s+\ep}}{|x-y|^{n+2s}}\,dy \\ &\asymp\int_0^1 \frac{r^{2s+\ep}}{r^{n+2s}}r^{n-1}\,dr \\ &=\int_0^1 r^{\ep-1}\,dr<\infty \end{aligned}$$ and $$\begin{aligned} I_2&:=\int_{\R^n\setminus B_x}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy \\ &\le c\int_{\R^n\setminus B_x}\frac{dy}{|x-y|^{n+2s}} \\ &\asymp\int_1^\infty\frac{r^{n-1}\,dr}{r^{n+2s}} \\ &=\int_1^\infty r^{-1-2s}\,dr<\infty. \end{aligned} $$ So, $$\sup_{x\in\Om}\int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy<\infty,$$ as desired.

On the other hand, if the extended function $u$ is not assumed to be Hölder-continuous on the entire space $\R^n$, then your desired conclusion will not hold in general. E.g., let $\Om$ be the open unit $\ell^\infty$-ball in $\R^n$ and let $$u(x):=\|x\|_\infty^{2s+\ep}\,1(x\in\Om)=\|x\|_\infty^{2s+\ep}\,1(\|x\|_\infty<1)$$ for all $x\in\R^n$. Then $u$ is Hölder-continuous on $\Om$ with exponent $2s+\ep$. However, for $x=(a,0,\dots,0)\in\Om$ with $a\uparrow1$ $$ \begin{aligned} \int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy &\ge\int_1^\infty dy_1\int_{\R^{n-1}}\, \frac{a^{2s+\ep}\,dy_2\cdots dy_n}{((y_1-a)^2+\sum_{j=2}^n y_j^2)^{n/2+s}} \\ &\to\int_1^\infty dy_1\int_{\R^{n-1}}\, \frac{dy_2\cdots dy_n}{((y_1-1)^2+\sum_{j=2}^n y_j^2)^{n/2+s}} \\ &=\frac12\,\int_{\R^n}\frac{dz}{|z|^{n+2s}} \\ &\asymp\int_0^\infty \frac{r^{n-1}\,dr}{r^{n+2s}}=\int_0^\infty r^{-1-2s}\,dr=\infty. \end{aligned} $$ So, here $\int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy$ is unbounded.

$\newcommand\ep\epsilon\newcommand\Om\Omega\newcommand\al\alpha\newcommand\R{\mathbb R}$According to your comment, $C^{0,2s+\ep}(\Om)$ is the set of all functions on $\Om$ that are Hölder-continuous with exponent $2s+\ep\in(0,1)$. It also appears that you extend the functions $u\in C^{0,2s+\ep}(\Omega)$ to $\R^n$ by letting $u:=0$ on $\R^n\setminus\Om$.

If such an extended function $u$ is Hölder-continuous on $\R^n$ with exponent $2s+\ep$, then there is some real $c>0$ such that $$|u(x)-u(y)|\le c|x-y|^{2s+\ep}$$ for all $x,y$ in $\R^n$, whence for $D$ defined as the diameter of $\Om$ and all $x\in\Om$ $$\int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy \ll\int_0^D \frac{r^{2s+\ep}}{r^{n+2s}}r^{n-1}\,dr=\int_0^D r^{\ep-1}\,dr<\infty$$ provided that $\ep>0$, as desired.

$\newcommand\ep\epsilon\newcommand\Om\Omega\newcommand\al\alpha\newcommand\R{\mathbb R}$According to your comment, $C^{0,2s+\ep}(\Om)$ is the set of all functions on $\Om$ that are Hölder-continuous with exponent $2s+\ep\in(0,1)$. It also appears that $\ep>0$ and that you extend the functions $u\in C^{0,2s+\ep}(\Omega)$ to $\R^n$ by letting $u:=0$ on $\R^n\setminus\Om$.

Take any such extended function $u$ that is Hölder-continuous on $\R^n$ with exponent $2s+\ep$. Taking into account the boundedness of $\Om$, we see that $$|u(x)-u(y)|\le\min(c,c|x-y|^{2s+\ep})$$ for some real $c>0$ and all $x,y$ in $\R^n$. Then for all $x\in\Om$, letting $B_x$ denote the (say open) unit ball in $\R^n$ centered at $x$, we have $$\int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy=I_1+I_2,$$ where $$I_1:=\int_{B_x}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy \le c\int_{B_x}\frac{|x-y|^{2s+\ep}}{|x-y|^{n+2s}}\,dy \asymp\int_0^1 \frac{r^{2s+\ep}}{r^{n+2s}}r^{n-1}\,dr=\int_0^1 r^{\ep-1}\,dr<\infty$$ and $$I_2:=\int_{\R^n\setminus B_x}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy \le c\int_{\R^n\setminus B_x}\frac{dy}{|x-y|^{n+2s}} \asymp\int_1^\infty\frac{r^{n-1}\,dr}{r^{n+2s}} =\int_1^\infty r^{-1-2s}\,dr<\infty.$$ So, $$\sup_{x\in\Om}\int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy<\infty,$$ as desired.

On the other hand, if the extended function $u$ is not assumed to be Hölder-continuous on the entire space $\R^n$, then your desired conclusion will not hold in general. E.g., let $\Om$ be the open unit $\ell^\infty$-ball in $\R^n$ and let $$u(x):=\|x\|_\infty^{2s+\ep}\,1(x\in\Om)=\|x\|_\infty^{2s+\ep}\,1(\|x\|_\infty<1)$$ for all $x\in\R^n$. Then $u$ is Hölder-continuous on $\Om$ with exponent $2s+\ep$. However, for $x=(a,0,\dots,0)\in\Om$ with $a\uparrow1$ $$ \begin{aligned} \int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy &\ge\int_1^\infty dy_1\int_{\R^{n-1}}\, \frac{a^{2s+\ep}\,dy_2\cdots dy_n}{((y_1-a)^2+\sum_{j=2}^n y_j^2)^{n/2+s}} \\ &\to\int_1^\infty dy_1\int_{\R^{n-1}}\, \frac{dy_2\cdots dy_n}{((y_1-1)^2+\sum_{j=2}^n y_j^2)^{n/2+s}} \\ &=\frac12\,\int_{\R^n}\frac{dz}{|z|^{n+2s}} \\ &\asymp\int_0^\infty \frac{r^{n-1}\,dr}{r^{n+2s}}=\int_0^\infty r^{-1-2s}\,dr=\infty. \end{aligned} $$ So, here $\int_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy$ is unbounded.