Let $P\neq \varnothing$ be a poset and $Q \subseteq P$ be both an order ideal and an order filter. Let $a,b \in P$ with $a\neq b$ and $a\in Q$.
Suppose there exists a fence in $P$ linking $a$ and $b$, that is, a sequence of elements $a_1, a_2, \ldots, a_{n-1}, a_n$ such that $a=a_1$, $a_n=b$, and $$a_1 > a_2 < a_3 > \cdots \# a_n,$$ where $\#$ is $<$ if $n$ is odd, and is $>$ if $n$ is even. Prove that $b \in Q$.
This is perhaps by induction on the size $n$ of the fence, being obvious for $n=1$. Then, suppose it is true for every $k<n$, that is, if $$a_1 > a_2 < a_3 > \cdots a_k$$ then $a_k \in Q$. Now what?
Thank you in advance
(Picking where you stopped.)
Suppose there is a $n$-size fence connecting $a_1=a$ and $a_n=b$.
Then there is a $n-1$-size fence connecting $a_1$ and $a_{n-1}$, and so, by the induction hypothesis, $a_{n-1} \in Q$.
Now, either $a_n>a_{n-1}$ or $a_n<a_{n-1}$ (depending on the parity of $n$).
In the former case, use the fact that $Q$ is an order-filter to conclude that $a_n \in Q$; in the later, use that $Q$ is an order-ideal.
So in any case, $b=a_n \in Q$.