In fact a substantially stronger embedding is true:

Theorem. If $n\geq 2$, $p>1$, and $\Omega\subset\mathbb{R}^n$ is a bounded domain with Lipschitz boundary, then $W^{1,p}(\partial\Omega)\subset W^{1-\frac{1}{q},q}(\partial\Omega)$, where $q=\frac{np}{n-1}$. That is, there is a bounded linear extension operator $$ E:W^{1,p}(\partial\Omega)\to W^{1,q}(\Omega)\cap C^\infty(\Omega). $$

When $p=2$ i.e., $W^{1,2}(\partial\Omega)=H^1(\partial\Omega)$, we have $q=\frac{2n}{n-1}>2$. Hence the extension is $$ E:H^1(\partial\Omega)\to W^{1,q}(\Omega)\subset H^1(\Omega), $$ proving (indirectly) that $H^1(\partial\Omega)\subset H^{\frac{1}{2}}(\partial\Omega)$ (because traces of $H^1(\Omega)$ are in $H^{\frac{1}{2}}(\partial\Omega)$.

The above theorem is known, but it is hard to find a reference to an elementary proof of it. I will add references later.

In fact a substantially stronger embedding is true:

Theorem. If $n\geq 2$, $p>1$, and $\Omega\subset\mathbb{R}^n$ is a bounded domain with Lipschitz boundary, then $W^{1,p}(\partial\Omega)\subset W^{1-\frac{1}{q},q}(\partial\Omega)$, where $q=\frac{np}{n-1}$. That is, there is a bounded linear extension operator $$ E:W^{1,p}(\partial\Omega)\to W^{1,q}(\Omega)\cap C^\infty(\Omega). $$

When $p=2$ i.e., $W^{1,2}(\partial\Omega)=H^1(\partial\Omega)$, we have $q=\frac{2n}{n-1}>2$. Hence the extension is $$ E:H^1(\partial\Omega)\to W^{1,q}(\Omega)\subset H^1(\Omega), $$ proving (indirectly) that $H^1(\partial\Omega)\subset H^{\frac{1}{2}}(\partial\Omega)$ (because traces of $H^1(\Omega)$ are in $H^{\frac{1}{2}}(\partial\Omega)$.

The above theorem is known. For a reference to a proof, see: https://mathoverflow.net/a/322635/121665.