The Hilbert distance is a genuine metric, with straight segments as geodesics

Symmetry and positivity. Write $[a,x,y,b]=\frac{|ay|}{|ax|}\cdot\frac{|bx|}{|by|}$. Since $a<x<y<b$ on the line, $|ay|=|ax|+|xy|>|ax|$ and $|bx|=|by|+|xy|>|by|$, so both factors exceed $1$ when $x\ne y$, hence $[a,x,y,b]>1$ and $d_\Omega(x,y)>0$; when $x=y$ both factors equal $1$. Relabelling $(x,y,a,b)\mapsto(y,x,b,a)$ sends $\frac{|ay||bx|}{|ax||by|}$ to $\frac{|bx||ay|}{|by||ax|}$, the identical number, so $d_\Omega(x,y)=d_\Omega(y,x)$.

Additivity on a line. Let $x,y,z$ be collinear with $y$ between $x$ and $z$, and let $a,b$ be the (common) boundary intersections of that line. Writing all four one-dimensional distances along the line, $$[a,x,y,b]\cdot[a,y,z,b]=\frac{|ay|}{|ax|}\frac{|bx|}{|by|}\cdot\frac{|az|}{|ay|}\frac{|by|}{|bz|}=\frac{|az|}{|ax|}\frac{|bx|}{|bz|}=[a,x,z,b],$$ since the factors $|ay|$ and $|by|$ cancel. Taking $\frac12\ln$ of both sides gives $d_\Omega(x,y)+d_\Omega(y,z)=d_\Omega(x,z)$, exactly the equality case of the triangle inequality, so the straight segment $[x,z]$ realizes the Hilbert distance and is a geodesic.

Triangle inequality off a line. For $x,y,z$ not collinear, convexity of $\Omega$ is what is needed: projecting the pair $(x,z)$ through $y$ only ever enlarges the relevant cross-ratio compared with going straight from $x$ to $z$, because the boundary points seen from $y$ lie no closer to $x$ or $z$ than the boundary points seen directly along $xz$ — this monotonicity, proved carefully using the convexity of $\Omega$, gives $d_\Omega(x,z)\le d_\Omega(x,y)+d_\Omega(y,z)$ in general (see Hilbert 1895; a full derivation is given in the Handbook of Hilbert Geometry, Ch. 1).