More formally, is there any example where the following is possible?
Given a linear subspace $\mathbf{A}$ and an element $\mathbf{x}\in\mathbf{A}$ there exists another linear subspace $\mathbf{B}$ such that $\mathbf{x}\notin\mathbf{A}\cap\mathbf{B}$, yet $\dim\mathbf{A}=\dim\mathbf{A}\cap\mathbf{B}$.
Intuitively this seems trivially impossible, but I'm having trouble seeing this clearly.
On
Not, if subspace is finite dimensional. For if you have a point $v\notin W $, &
let $W=$ span{$v_1,v_2,....,v_n$} where $v_1,...,v_n$ are basis of $W$, then it can be shown that $v, v_1,v_2,...,v_n$ are linearly independent and you might already know that maximal set of linear independent vectors constitute a basis.
This cannot happen in finite dimension because $$ \dim (A) = \dim (A\cap B) $$ and $$ A \supseteq A\cap B $$ implies $A=A\cap B$. Since $x \in A$, we cannot have $x \notin A\cap B$.