This is a counterexample to Vopenka's principle phrased slightly differently: as "in any proper class of first-order structures, one elementarily embeds into the other."
Working in $V=L$, I claim that $\{L_{\kappa^+}: \kappa\in Card\}$ is a counterexample to Vopenka's principle.
Suppose $\kappa<\lambda$ are cardinals, and $L_{\kappa^+}$ elementarily embeds into $L_{\lambda^+}$, via $j$. Let $\mathcal{U}$ be the set of subsets $X$ of $\kappa$ such that $\kappa\in j(X)$. Clearly $\mathcal{U}$ is an ultrafilter; I claim $\mathcal{U}$ is countably closed. Let $S=\{X_i: i\in\omega\}$ be a sequence of subsets of $\kappa$ such that $X_i\in\mathcal{U}$ for every $i\in\omega$. Note that since $V=L$, the sequence $S$ exists in $L_{\kappa^+}$, and so we can look at $j(S)$. Note first of all that $j(S)$ is an $\omega$-sequence, whose terms are exactly the $j(X_i)$, since $\omega$ can't be moved by $j$. We have $$\forall Y\in j(S), \kappa\in Y,$$ that is, $$\kappa\in\bigcap j(S).$$ But since $S\in L_{\kappa^+}$ and $j$ is elementary, $\bigcap j(S)=j(\bigcap(S))$, so we are done.
So $\mathcal{U}$ is a countably complete ultrafilter on $\kappa$. But this is equivalent to $\kappa$ being measurable, and $V=L$ implies measurable cardinals don't exist.