Inclusive Disjunction Problem

Given $(P:=3)$ and $x$ is an arbitrary element of the interval $S:= [1,\,5]$ except $3$, what would it be as $P \in S$?

Based on the given information above, verify that $(P < x) \lor (P > x)$, where $x$ is any arbitrary element except $3$ in $S$ and supposing $\lor$ is the inclusive disjunction?

This is my scratch work of proving:

Define $J := P < x,\space K := P > x,\space x≠3$.

1. Take any $x \in[A,P]$ or
2. Take any $x \in [P,B]$

Since from 1. $\left(J=\mathbf{TRUE}\right)\,\lor\,> \left(K=\mathbf{FALSE}\right)=\mathbf{TRUE}$, and from 2.$\left(J=\mathbf{FALSE}\right)\,\lor\,> \left(K=\mathbf{TRUUE}\right)=\mathbf{TRUE}$. Thus the statement istrue.

If there's any, I reckon there is, could you please kindly to point out the mistake and why it is wrong?

Thank you in advance.