Showing $H^1(\partial\Omega) \subset H^{\frac 12}(\partial\Omega)$ is continuous?

Lions and Magenes define $H^\frac12$ by complex interpolation, hence your desired property holds by assumption; but they always assume $\Omega$ to have smooth boundary.

In the books of Adams and Grisvard you will find some related results, but - as far as I have (quickly) seen - not exactly what you look like.

Here is a direct argument based on the definition by the Gagliardo norm $$ \Vert u \Vert_{H^{1/2}}^2 = \int_{\mathbb{R}^N}\int_{\mathbb{R}^N} \frac{\vert u (x) - u (y)\vert^2}{ \vert x - y \vert^{N + 1}}\,dx\,dy + \int_{\mathbb{R}^N} \vert u \vert^2. $$

In a first step you apply a weighted Hardy inequality on the ball $B_1 (y)$ to write $$ \begin{split} \int_{\mathbb{R}^N} \frac{\vert u (x) - u (y)\vert^2}{ \vert x - y \vert^{N + 1}}\,dx\,dy &= \int_{B_1 (y)} \frac{\vert u (x) - u (y)\vert^2}{ \vert x - y \vert^{N + 1}}\,dx\,dy + \int_{\mathbb{R}^N \setminus B_1 (y)} \frac{\vert u (x) - u (y)\vert^2}{ \vert x - y \vert^{N + 1}}\,dx\,dy\\ &\le \int_{B_1 (y)} \frac{\vert \nabla u (x)\vert^2}{\vert x - y \vert^{N - 1}} + \vert{u(x)}\vert^2\,dx + 2 \int_{\mathbb{R}^{N }\setminus B_1 (y)} \frac{\vert u (x) \vert^2}{\vert x - y\vert^{N + 1}} \,dx + 2 \vert u (y)\vert^2 \int_{\mathbb{R}^{N} \setminus B_1} \frac{1}{\vert z\vert^{N + 1}} \,dx. \end{split} $$ The conclusion comes by integrating with respect to $y \in \mathbb{R}^N$ and then applying Fubini's theorem and using the facts that $$ \int_{B_1} \frac{1}{\lvert z \rvert^{N - 1}} \,dz < \infty \text{ and }\int_{\mathbb{R}^N \setminus B_1} \frac{1}{\lvert z \rvert^{N + 1}} \,dz < \infty . $$

To cover a smooth domain, the function can be extended to the whole space $\mathbb{R}^N$.

This argument also works for the spaces $W^{s, p}$ and $W^{1, p}$, where $$ \Vert u \Vert_{W^{s, p}}^p = \int_{\mathbb{R}^N}\int_{\mathbb{R}^N} \frac{\vert u (x) - u (y)\vert^p}{ \vert x - y \vert^{N + sp}}\,dx\,dy + \int_{\mathbb{R}^N} \vert u \vert^p. $$