The title might be a little imprecise: The question is whether $H^{1/2}(0,2)$ is a closed subspace of the space $X$ of functions $f$ that are on the form $$f= \begin{cases} g & \text{in $(0,1)$}\\ h & \text{in $(1,2)$} \end{cases}, \qquad g \in H^{1/2}(0,1), h\in H^{1/2}(1,2)$$ i.e. I glue two $H^{1/2}$ functions together. The norm on $X$ is the $H^{1/2}(0,1)\times H^{1/2}(1,2)$ norm.
Let $f_n$ be a sequence of functions in $H^{1/2}(0,2)$, i.e. $$\|f_n\|_{H^{1/2}(0,2)}^2=\int_0^2 \int_0^2 \frac{|f_n(x)-f_n(y)|^2}{|x-y|^2} \, dx\, dy < \infty$$ (denoting $g_n = \mathbb{1}_{(0,1)}f_n$ and $h_n = \mathbb{1}_{(1,2)}f_n$) such that there exists $f\in X$ $$\lim_{n\rightarrow\infty}\|f_n-f\|_X=\lim_{n\rightarrow\infty}(\|g_n-g\|_{H^{1/2}(0,1)}+\|h_n-h\|_{H^{1/2}(1,2)})=0.$$ Can we conclude that $f$ is in $H^{1/2}(0,2)$?
Manipulating the $H^{1/2}(0,2)$ norm with triangle inequalities and so on I reduced the problem to showing $$\int_{0}^1 \frac{(g_n(x)-g(x))^2}{1-x}\, dx < \infty,$$ and $$\int_{1}^2 \frac{(h_n(x)-h(x))^2}{y-1}\, dy < \infty,$$ for large enough $n$. However this is not clear to me.