For the sake of completeness, an expansion on the comment by Mike Miller:

In Evans/Gariepy, Thm. 4.3.1, it is proven that if the boundary of $\Omega$ is Lipschitz, then for $1 \leq p < \infty$ there is a continuous linear trace operator $T$ from $W^{1,p}(\Omega)$ to $L^p(\Omega;\mathcal{H}_{n-1})$ which satisfies $$Tf = f \quad \text{on $\partial \Omega$}$$ if $f \in W^{1,p}(\Omega) \cap C(\overline\Omega)$.

Observing that $W^{1,p}_0(\Omega)$ is a closed subspace of $W^{1,p}(\Omega)$ and taking $g \in W^{1,p}_0(\Omega)$ and a sequence $(g_k) \subset C_c^\infty(\Omega)$ such that $g_k \to g$ in the $W^{1,p}(\Omega)$ norm, we obtain $$Tg = \lim_k Tg_k = g_k = 0 \quad \text{in $L^p(\partial\Omega,\mathcal{H}_{n-1})$}.$$ Thus, if $g \in W^{1,p}_0(\Omega) \cap C(\overline\Omega)$, then $0 = Tg = g$ everywhere on $\partial\Omega$.

It is maybe worthwile to note (Thm. 5.3.2 in Evans/Gariepy) that $T$ is in fact given by $$Tf(x) := \lim_{r\searrow0}\frac1{|B_r(x) \cap \Omega|} \int_{B_r(x) \cap \Omega} f$$
which can be a useful thing to look at also in situations where $\partial\Omega$ is less regular.

*Evans, Lawrence C.; Gariepy, Ronald F.*, Measure theory and fine properties of functions, Studies in Advanced Mathematics. Boca Raton: CRC Press. viii, 268 p. (1992). ZBL0804.28001.