A counterexample is $\omega+1$, turned upside down. However, it is true that every infinite distributive lattice contains either a non-principal prime ideal or a non-principal prime filter. This is due to the following argument:

Any infinite distibutive lattice contains at least a non-principal ideal or a non-principal filter (not necessarily prime). We may assume that $J$ is a non-principal ideal, so that $j^* = \bigvee J \notin J$. Let $G = \{y\in L: y \geq j^*\}$ be the principal filter generated by $j^*$. As $J\cap G = \emptyset$ we can use the Prime Ideal Theorem and get a prime ideal $P$ such that $J\subseteq P$ and $P\cap G = \emptyset$, which implies $j^* \notin P$.

Next we show that $P$ is not principal: if we had $p^*:=\bigvee P \in P$ then $J\subseteq P$ would imply $p^* \geq j^* = \bigvee J$ and $j^* \in P$ because $P$ is a down-set. But by construction $j^* \notin P$. Therefore $P$ is a non-principal prime ideal.

A counterexample is $\omega+1$, turned upside down. However, it is true that every infinite distributive lattice contains either a non-principal prime ideal or a non-principal prime filter. This is due to the following argument:

Any infinite distributive lattice contains at least a non-principal ideal or a non-principal filter (not necessarily prime). We may assume that $J$ is a non-principal ideal, so that $j^* = \bigvee J \notin J$. Let $G = \{y\in L: y \geq j^*\}$ be the principal filter generated by $j^*$. As $J\cap G = \emptyset$ we can use the Prime Ideal Theorem and get a prime ideal $P$ such that $J\subseteq P$ and $P\cap G = \emptyset$, which implies $j^* \notin P$.

Next we show that $P$ is not principal: if we had $p^*:=\bigvee P \in P$ then $J\subseteq P$ would imply $p^* \geq j^* = \bigvee J$ and $j^* \in P$ because $P$ is a down-set. But by construction $j^* \notin P$. Therefore $P$ is a non-principal prime ideal.