Separation Axioms: is it true that $T_4 \Rightarrow T_3 \Rightarrow T_2 \Rightarrow T_1 \Rightarrow T_0$?

I personally prefer that the $T_i$ axioms form a decreasing hierarchy as you suggest. $T_2 \implies T_1 \implies T_0$ already holds and it makes sense to extend this into the "higher" ones.

This leaves the "word"-versions (normal, regular, completely regular, corresponding to $T_{3\frac12}$ inbetween $T_4$ and $T_3$) for the "pure" point and closed set and closed sets versions of the axioms.

But having normal but no (or hardly any) closed sets to separate (like for the indiscrete space) makes it a rather "empty" notion I think. Normal is useful as it's precisely what needed for Urysohn's lemma, so it's handy to have a separate term for that. Adding $T_1$ (or $T_0$) to normal and regular at least leaves us with some closed sets (singletons etc) to separate. It makes the property more "interesting" and achieves the desired implications I mentioned.

There are similar debates among topology texts whether to include Hausdorffness in "compact" or "paracompact" etc. (see "quasicompact").

It's good to be aware that people don't agree on terms and always look in the introductions of papers (or indexes of books) to see how a symbol $T_3$ or word like "regular" is used. There are different traditions and if you're not careful you can misquote (or wrongly use) a paper's result.

Make a personal choice and stick to it would be my advice.

From Munkres’ Topology, section 31:

A two-point space in the indiscrete topology satisfies the other parts of the definitions of regularity and normality, even though it is not Hausdorff.

There is a minor deviation in terminologies: Some authors adopt the notion of $𝑇_3$ (or regular) being “a point and a closed set (not including that point) can be separated”, while others write $𝑇_3$ to say the bold quoted sentence plus Hausdorff. Similar for $T_4$ and its synonym normal.

In terms of your lecture, a two-point space in the indiscrete topology is $T_3$ and $T_4$, but not $T_2$. Thus $T_3\Rightarrow T_2$ is not valid.

P.S. We have $T_3+T_1\Rightarrow T_2$ and $T_4+T_1\Rightarrow T_2$, so this example does not satisfy $T_1$ as well.