I think there's a mistake in the proof that \$$\mathcal{X}^\text{op}\$$ is a \$$\mathcal{V}\$$-enriched category:

> b) We also need to check that for all objects \$$x,x',x'' \$$ of \$$\mathcal{X}^{\text{op}} \$$ we have
>
> $\mathcal{X}^{\text{op}} (x,x') \otimes \mathcal{X}^{\text{op}}(x',x'') \le \mathcal{X}^{\text{op}}(x,x'') .$
>
> Using the definitions, this just says that for all objects \$$x,x',x'' \$$ of \$$\mathcal{X}\$$ we have
>
> $\mathcal{X}(x'',x') \otimes \mathcal{X}(x',x) \le \mathcal{X}(x'',x) .$
>
> This is true because \$$\mathcal{X}\$$ is an enriched category. \$$\qquad \blacksquare \$$.
>
> I just learned something: I thought we needed \$$\mathcal{V}\$$ to be commutative here, but apparently we don't.

Surely the definitions tell us that

$\mathcal{X}^{\text{op}} (x,x') \otimes \mathcal{X}^{\text{op}}(x',x'') = \mathcal{X}(x',x) \otimes \mathcal{X}(x'',x')$

So we _do_ need \$$\mathcal{V}\$$ to be commutative to switch this to \$$\mathcal{X}(x'',x') \otimes \mathcal{X}(x',x)\$$.

In the absence of commutativity/symmetry, all we can prove is that \$$\mathcal{X}^\text{op}\$$ is a \$$\mathcal{V}^\text{op}\$$-enriched category, where \$$\mathcal{V}^\text{op}\$$ is \$$\mathcal{V}\$$ with the same partial order but the opposite monoidal multiplication, ie \$$u \otimes_{\mathcal{V}^\text{op}} v = v \otimes_\mathcal{V} u\$$.