Solution to problem 1 in the $s=t=1$ case
In the $(1,1)$ case, change the pseudometric to $(x,y)\mapsto |xy|$ instead of $(x,y)\mapsto |x^2 - y^2|$. This is merely a change of coordinates.
A function $f:\mathbb R^{1,1}\to \mathbb R^{1,1}$ is "pseudocontinuous" if and only if:
$$f(x,y)\equiv(g(x),h(y)),$$ or $$f(x,y)\equiv(h(y),g(x)),$$ for any pair of continuous functions $f,g:\mathbb R\to\mathbb R$.
Why? I list the sequence of lemmas which one should prove:
Lemma 1: A "cross" (that is, the union of a line parallel to the $x$ axis with a line parallel to the $y$ axis) necessarily gets mapped to another cross by any pseudocontinuous function $f$.
Lemma 2: A horizontal line or vertical line gets mapped to a horizontal or vertical line by pseudocontinuous $f$.
Lemma 3: For fixed $x$, the function $y \mapsto f(x,y)$ is continuous. The same is true for $x \mapsto f(x,y)$ for fixed $y$.
Lemma 4: The function $f$ is continuous everywhere.
Theorem: The above statement.