What are some examples of consistent theories $T_i$ (extending elementary arithmetic EA) such that for $∀i∈ℕ \,\, T_i ⊢ \mathrm{Con}(T_{i+1})$?

Such theories exist; see for example An infinitely descending sequence of sound theories each proving the next consistent. However, in the link, the theories are given using (non-well-founded) self-reference, which in an intuitive informal sense, leaves one wondering what exactly is each theory asserting.

Thus, I am especially interested in natural or intuitive examples that do not use self-reference. Here is what I have.

If we only require $\mathrm{PA} ⊬ (\mathrm{Con}(T_{i+1}) ⇒ \mathrm{Con}(T_i))$, then for an example, fix a proof system that is constructively provably in $\mathrm{EA}$ polynomial time equivalent to sequent calculus. Let $n$ (provably in EA) be the number of bits in the shortest inconsistency in $\mathrm{PA}$ if there is any. If $T_i = \mathrm{EA} + \mathrm{Con}(\mathrm{PA}) ∨ 2^i ∤ ⌊\log \log \log n⌋$, then for all $i$, $T_i ⊢ T_{i+1}$ and $\mathrm{PA} ⊬ (\mathrm{Con}(T_{i+1}) ⇒ \mathrm{Con}(T_i))$. (To get the unpredictability of $n$, the proof relies on $\mathrm{PA}$ provability of $\mathrm{PA}$ bounded consistency in polynomial time, and the impossibility of a subpolynomial proof length of bounded consistency.)

To get $T_i ⊢ \mathrm{Con}(T_{i+1})$, I have two potential natural examples, but I do not have a proof that they work.

*Update:* Per Fedor Pakhomov's answer, the original constructions (given in brackets for reference) did not work; below are the modified versions. See the answers for constructions that have been shown to work; the answers complement each other.

*Construction 1:* Let $S_0 = \mathrm{PA}$ and $S_{i+1} = \mathrm{PA} + \mathrm{Con}(S_i)$, and let $h_{ε_0}$ be the $ε_0$ function in the Hardy hierarchy.
Set $T_i = \mathrm{PA} + ∀n \,\, (h_{ε_0 2}(i+n) \text{ exists} ⇒ \mathrm{Con}(S_n))$ [the original used $h_{ε_0}$, but perhaps even $h_{ε_0 2} = λx.h_{ε_0}(h_{ε_0}(x))$ is too small]

*Construction 2:* Let $S_0 = \mathrm{PA}$ and $S_{i+1} = \mathrm{PA} + \mathrm{Con}(S_i)$. We can arrange that $⌈\mathrm{Con}(S_i)⌉$ is $i^{O(1)}$, but we only need $2^{2^{2^{O(i)}}}$. Let $\mathrm{Con}_m(A)$ mean that $A$ has no inconsistency shorter than $m$ bits.

Set $T_i = \mathrm{PA} + ∀n \, (\mathrm{Con}_{2^{n^i}}(S_{n+i}) ⇒ \mathrm{Con}(S_n))$ [the original used $S_n$ in place of $S_{n+i}$].

*Note:* If we can show in $\mathrm{PA}$ that each $T_i$ (for $i>0$) is consistent relative to $S=∪S_i$, then the construction works because if $S$ is inconsistent, then working in $T_i$, $T_{i+1}$ is equivalent to $S_{i'}$ for a low enough $i'$ so as to be consistent.

Also, the well-ordering of $Π^0_1$ ordinals imposes fundamental limits on descending consistency chains. I suspect that for sound $T$, $∃i \, T_i ⊬ ∀j \, (T_j ⊢ \mathrm{Con}(T_{j+1}))$.