Let $\ a_3\ $ be the maximal area of convex sets
$\ X\subseteq \mathbb I^2\setminus T\ $
for the worst $3$-element subset $\ T\subseteq\mathbb I^2,\ $ i.e. for $\ T\ $ which leads to the minimal said maximum. We saw in my earlier answer that $\ a_3\ge\frac 13.\ $ My whole ambition here is to improve upon this bound (by *my ambition* I mean that I do not attempt to
squeeze a still stronger result from my or any other method). The numerical result is stated at the very end of this text.

Thus, let $\ |T|=3.\ $ If the center of the square $\mathbb I^2$ does not belong to the interior of the triangle $\ \triangle(T)\ $ then the said maximal convex area outside $T$ is at least $\ \frac 12$.

From now on I assume that the center belongs to the interior of $\ \triangle(T)\ $ (this means that $T$ is not collinear). The rest of the argument consists of two parts plus a conclusion.

**Part 1.** Let $\ t\ $ be the area of $\ \triangle(T).\ $ Then
the radius $\ r\ $of the circle inscribed into the triangle satisfies
inequality:

$$ r\ \le\ 3^\frac {-3}4 \cdot t $$

Furthermore, the center of the square is between one of the straight
lines of $\ T,\ $ call it $L$, and the line parallel to $L$ which
passes through the center of the inscribed circle. Let straight
line $\ Ł \ $ be parallel to $\ L,\ $ and apart from the center of the square by $r$, and such that $L$ is between the center of the square and $Ł$.

We see that the (convex) side of the square outside of $\ Ł,\ $ and away from from the said centers has area $\ \le\ \alpha := \frac 12 - r,\ $

(**EDIT** I've just fixed the direction of this above inequality for area)

i.e.

$$ \alpha\,\ \ge\,\ \frac 12\ -\ 3^\frac {-3}4 \cdot t $$

This way we have obtained a convex set of area $\alpha$ which is disjoint with the interior of $\triangle(T)$.

**Part 2:** The three straight lines which pass through the pairs of points of $T$ form some kind of a *Venn diagram* (not really) which consists of $\triangle T$, of three angular areas $\ K_1\ K_2\ K_3,\ $
and of three infinite triangles (to me they look like infinite trapezoids, but never mind) $\ B_1\ B_2\ B_3.\ $ The enumeration is such that
$\ K_i\cap B_i=\emptyset\ $ for each $\ i=1\ 2\ 3$.

Every union

$$ S_{ijm}\ :=\ K_i\cup B_j\cup K_m $$

combines to one side of the said straight lines, which is disjoint with the interior of $\ \triangle(T),\ $ whenever $\ i+j+m = 6\ $ and
$\ |\{i\ \ j\,\ m\}| = 3.\ $

Let $\ \sigma\ $ be the sum of the areas of $S_{123}$ and $S_{231}$ and
$S_{312}.\ $ Let $\ \kappa\ $ be the area of $\ K_1\cup K_2\cup K_3.\ $
Let $\ \gamma\ $ be the sum of areas of $\ \triangle(T)\cup B_1\ $ and
$\ \triangle(T)\cup B_2\ $ and $\ \triangle(T)\cup B_3.\ $ Then:

$$ 2\cdot\sigma + \gamma\ =\ 3 + 2\cdot t + \kappa $$
hence
$$ 2\cdot\sigma + \gamma\ \ge\ 3 + 2\cdot t $$

The left-hand side expression is the sum of $9$ convex sets which have pair-wise disjoint interiors, and which have their interiors disjoint from $T$--these convex sets are of the form $S_{ijm}$ and $K_i\cup\triangle(T)$.
Thus, at least one of these $9$ sets has area

$$ \beta\ \ge \frac 13 + \frac 29\cdot t $$

**Conclusion** The above estimates for $\alpha$ and $\beta$ are a decreasing and increasing function respectively. They are equal at

$$ t_0\ :=\ \frac 16\cdot\left(\frac 29 + 3^\frac{-3}4\right)^{-1} $$

By considering the case $\ t\le t_0$, then $\ t\ge t_0\ $ respectively,
we obtain:

**Theorem**
$$ a_3\,\ \ge\,\ \frac 12\ -\ 3^\frac{-3}4 \cdot t_0\,\ =
\,\ \frac 13\ +\ \frac 29\cdot t_0 $$

In case you are curious about the numerical value of the just obtained
lower bound on $a_3$, it is:
$$ a_3\ \ge\ 0.38937\ldots $$
(I am glad that it is clearly greater than $\ \frac 13$).