Scott, your snake equation proof looks good, except that you only show that the two profunctors are equal along the diagonal. That is, you only show that for all \$$x\$$,
$(1_X \times \cup_X ) (\cap_X \times 1_X )((1,x),(x,1)) = 1_X(x,x)$
You need to show that for all \$$x,\tilde{x}\$$,
$(1_X \times \cup_X ) (\cap_X \times 1_X )((1,x),(\tilde{x},1)) = 1_X(x,\tilde{x}).$
Replacing \$$x\$$ by \$$\tilde{x}\$$ in the appropriate places in your proof, you get that
\$$(1_X \times \cup_X ) (\cap_X \times 1_X )((1,x),(\tilde{x},1))\$$
if and only if there exist \$$x'', x^*, x'\$$ such that \$$x\le x'' \le x^* \le x' \le \tilde{x}\$$,
and of course this hold if and only if \$$x\le\tilde{x}\$$ or in other words \$$1_X(x,\tilde{x})\$$.

On a stylistic note, I think 0 is better than 1 as the name for the only element in \$$\mathbf{1}\$$, as John used in the last lecture. I've seen others here use a bullet or star symbol, which I think are also better than 1.