If $\,f(x,y)\,$ is a polynomial with integer coefficients then by the polynomial congruence rule
$$\bmod 2\!:\,\ \ f(m,n)\,\equiv\, f(\overbrace{m\bmod 2}^{\large \overline{m}},\, \overbrace{n\bmod 2}^{\large\overline{n}})\qquad $$
So $\,f(m,n) = 0\,\Rightarrow\, f(\bar m,\bar n) \equiv 0\pmod{\!2}\ $ so contra+, $\ f(\bar m,\bar n) \not\equiv 0\pmod{\!2}\,\Rightarrow\, f(m,n)\neq 0$
i.e. any root $\,(m,n)\,$ of $f$ persists as a root $\!\bmod 2.\,$ OP is the special case when $f$ is the determinant function - which is indeed a polynomial function of its entries (with integer coefficients). Concretely
$$d = \left|\begin{array}{}\color{#c00}{11} & \color{#0a0}{22}\\ \color{#0a0}{33} & \color{#c00}{55}\end{array}\right| = \left\{\begin{align}&f(11,22,33,55) = \color{#c00}{11(55)}\!-\!\color{#0a0}{22(33)}\\
\ \equiv\ &f(\ \ 1,\ \ 0,\ \ 1,\ \ 1) \equiv \color{#c00}{\ \ 1(\ \ 1)}-\color{#0a0}{0(\ \ 1)}\equiv 1\!\!\!\pmod{\!2}\end{align}\right.\qquad$$
therefore $\,d\equiv 1\pmod{\!2},\,$ i.e. $\,d\,$ is odd, so $\,d\neq 0.$
Remark $ $ The same works for any modulus $\,m,\,$ e.g. we could use $\,m = 9\,$ ("casting out nines") exactly as above (or more generally as a check of any such polynomial computation), or we could compare unit digits $\bmod m\!=\!10\!:\ \,d\equiv \color{#c00}{1(5)}-\color{#0a0}{2(3)}\equiv -1\ $ so $\,d\neq 0.$
For motivation you might find instructive this post on the parity root test for polynomials. Such parity arguments are a prototypical way of solving ring (algebraic) problems via information gleaned from (simpler) modular images (an algebraic way of "dividing and conquering").