This is problem 2.4 from "Numerical Solution of Partial Differential Equations by the Finite Element Method" by Claes Johnson.
Let $\Omega$ be a square with boundary $\Gamma$. Show that there is a constant C such that $$\left(\int_\Gamma v^2 ds\right)^{1/2}\le C||v||_{H^1(\Omega)}, \forall v\in H^1(\Omega)$$ Then use that result to show that $$L(v)=\int_\Gamma gv ds$$ is continuous if $g\in L^2(\Gamma)$.
Idea: I think I need to use Green's theorem, and the fact that were are specifically working with a square. However, I don't know how to turn that into an actual mathematical argument.
The first inequality is indeed Trace theorem (see Evans PDE page 272). The idea is as follows: Suppose $\Omega=[0,1]\times[0,1]$, and $\Gamma = \Gamma_1\cup \Gamma_2\cup \Gamma_3\cup \Gamma_4$, where \begin{align*} \Gamma_1&=[0,1]\times \left\{ 0 \right\};\\ \Gamma_2&=[0,1]\times \left\{ 1 \right\};\\ \Gamma_3& = \left\{ 0 \right\}\times[0,1];\\ \Gamma_4& = \left\{ 1 \right\}\times[0,1]. \end{align*} Then select a cut off function $\zeta\in C^{\infty}(\Omega)$ s.t. $\zeta=1$ on $\Gamma_1$ and $\zeta=0$ on $\Gamma_2$. \begin{align*} \int_{\Gamma_1}v^2ds &= -\int_{\Omega}(\zeta v^2)_{x_2}dx_1dx_2\\ & =-\int_{\Omega}\zeta_{x_2}v^2+\zeta vv_{x_2}dx_1dx_2\\ & \leq C\int_{\Omega}|v|^2+|Dv|^2dx_1dx_2 \end{align*} where we used Young's inequality. In a similar way, one can show that this inequality holds for $\Gamma_2$, $\Gamma_3$ and $\Gamma_4$. In all, \begin{equation} \left( \int_{\Gamma}v^2ds \right)^{\frac{1}{2}}\leq C\|v\|_{H^{1}(\Omega)}. \end{equation} For the second part, since $L$ is a linear funciotnal over $H^{1}(\Omega)$, therefore it is continuous if and only if it is bounded, that is \begin{equation} L(v)=\int_{\Gamma}^{}gvds\leq \|g\|_{L^2(\Gamma)}\|v\|_{L^2(\Gamma)}\leq C\|v\|_{H^1(\Omega)}. \end{equation}