[proofplan]
We identify simplices of the order complex $\triangle(P)$ with chains in the poset $P$. A simplex lies in the link of $x$ exactly when it does not contain $x$ and becomes a chain after adjoining $x$. Such a chain splits uniquely into the part below $x$ and the part above $x$, and conversely any lower chain and upper chain combine with $x$ to form a chain. This is precisely the defining simplex condition for the simplicial join $\triangle(P_{<x})*\triangle(P_{>x})$.
[/proofplan]
[step:Characterize simplices in the link of $x$ as chains compatible with adjoining $x$]
Let $\sigma\subset P$ be a simplex of $\triangle(P)$. By definition of the order complex, this means that $\sigma$ is a chain in $P$. By definition of the link of the vertex $x$, the simplex $\sigma$ lies in $\operatorname{lk}_{\triangle(P)}(x)$ if and only if
\begin{align*}
x\notin \sigma \quad \text{and} \quad \sigma\cup\{x\}\in \triangle(P).
\end{align*}
Equivalently, $\sigma$ is a chain in $P$ not containing $x$, and $\sigma\cup\{x\}$ is also a chain in $P$.
[/step]
[step:Split every link simplex into its lower and upper parts]
Assume $\sigma\in \operatorname{lk}_{\triangle(P)}(x)$. Define
\begin{align*}
\sigma_{<}:=\sigma\cap P_{<x}, \qquad \sigma_{>}:=\sigma\cap P_{>x}.
\end{align*}
For every $y\in \sigma$, the set $\sigma\cup\{x\}$ is a chain, so $y$ is comparable with $x$. Since $x\notin\sigma$, either $y<x$ or $y>x$. Hence
\begin{align*}
\sigma=\sigma_{<}\cup\sigma_{>},
\end{align*}
and the union is disjoint. Because $\sigma$ is a chain, the subsets $\sigma_{<}$ and $\sigma_{>}$ are chains in the induced subposets $P_{<x}$ and $P_{>x}$, respectively. Therefore
\begin{align*}
\sigma_{<}\in \triangle(P_{<x}) \quad \text{and} \quad \sigma_{>}\in \triangle(P_{>x}).
\end{align*}
[guided]
Let $\sigma$ be a simplex in the link of $x$. The link condition has two parts: first, $\sigma$ itself is a simplex of $\triangle(P)$ and does not contain $x$; second, adjoining $x$ still gives a simplex of $\triangle(P)$. Since simplices of $\triangle(P)$ are exactly chains in $P$, the set $\sigma\cup\{x\}$ is a chain.
Now take any element $y\in\sigma$. Because $\sigma\cup\{x\}$ is a chain, the two elements $y$ and $x$ are comparable in $P$. Since $y\ne x$, exactly one of the two strict alternatives holds:
\begin{align*}
y<x \quad \text{or} \quad y>x.
\end{align*}
Thus every element of $\sigma$ lies either in $P_{<x}$ or in $P_{>x}$. We define the two parts
\begin{align*}
\sigma_{<}:=\sigma\cap P_{<x}, \qquad \sigma_{>}:=\sigma\cap P_{>x}.
\end{align*}
These two subsets are disjoint because no element can be both strictly below and strictly above $x$. They cover $\sigma$ because every element of $\sigma$ is comparable with $x$ and is not equal to $x$. Hence
\begin{align*}
\sigma=\sigma_{<}\cup\sigma_{>}.
\end{align*}
It remains to check that these two pieces are simplices in the two smaller order complexes. The set $\sigma_{<}$ is a subset of the chain $\sigma$, so it is itself a chain; all of its elements lie in $P_{<x}$ by definition. Therefore $\sigma_{<}\in\triangle(P_{<x})$. The same argument gives $\sigma_{>}\in\triangle(P_{>x})$. This proves that every simplex in the link decomposes into a lower simplex and an upper simplex.
[/guided]
[/step]
[step:Combine any lower chain and upper chain into a link simplex]
Conversely, let $\tau\in\triangle(P_{<x})$ and let $\upsilon\in\triangle(P_{>x})$. Define
\begin{align*}
\sigma:=\tau\cup\upsilon.
\end{align*}
The set $\tau$ is a chain in $P_{<x}$ and $\upsilon$ is a chain in $P_{>x}$. If $a\in\tau$ and $b\in\upsilon$, then $a<x$ and $x<b$, so transitivity gives $a<b$. Hence every two elements of $\sigma$ are comparable, and $\sigma$ is a chain in $P$. Also $x\notin\sigma$, since no element of $P_{<x}$ or $P_{>x}$ is equal to $x$. Finally, the same comparison shows that $\tau\cup\{x\}\cup\upsilon$ is a chain in $P$. Thus
\begin{align*}
\sigma\in\operatorname{lk}_{\triangle(P)}(x).
\end{align*}
[/step]
[step:Identify the resulting simplex set with the simplicial join]
By definition of the simplicial join, the simplices of $\triangle(P_{<x})*\triangle(P_{>x})$ are exactly the sets
\begin{align*}
\tau\cup\upsilon
\end{align*}
where $\tau\in\triangle(P_{<x})$ and $\upsilon\in\triangle(P_{>x})$. The previous two steps show that these are exactly the simplices of $\operatorname{lk}_{\triangle(P)}(x)$. The empty lower or upper part is allowed because the order complex of an empty poset contains the empty simplex, so the argument also covers the cases $P_{<x}=\varnothing$ or $P_{>x}=\varnothing$. Therefore
\begin{align*}
\operatorname{lk}_{\triangle(P)}(x)=\triangle(P_{<x})*\triangle(P_{>x}).
\end{align*}
[/step]
