FALSE: in every ordered field a closed bounded set is compact, so Heine-Borel needs no completeness

$S$ is nonempty and bounded: $1 \in S$ because $1 \ge 0$ and $1^2 = 1 < 2$ by [L5]; and every $q \in S$ satisfies $0 \le q < 2$, since $q \ge 2 \ge 0$ would give $q^2 \ge 2^2 = 4 > 2$ by [L4] and [L5], contradicting $q^2 < 2$.

$S$ has no greatest element: let $q \in S$ and put $h := \min\{\, 1,\ (2 - q^2)(2q+2)^{-1} \,\}$, a definition by cases on the total order of $\mathbb{Q}$; here $2q + 2 > 0$ because $q \ge 0$, and $2 - q^2 > 0$, so both entries are positive and $h > 0$ with $h \le 1$. Put $r := q + h$, so $r > q \ge 0$. Then $h^2 \le h$ because $0 < h \le 1$, and $h(2q+1) \le (2-q^2)(2q+1)(2q+2)^{-1} < 2 - q^2$ because $(2q+1)(2q+2)^{-1} < 1$ and $2 - q^2 > 0$; hence $r^2 = q^2 + 2qh + h^2 \le q^2 + h(2q+1) < q^2 + (2 - q^2) = 2$, so $r \in S$ and $q < r$.

$S$ is closed in $\mathbb{Q}$: let $q \in \mathbb{Q} \setminus S$, so $q < 0$, or $q \ge 0$ and $q^2 \ge 2$, in which case $q^2 \ne 2$ by [L3] gives $q^2 > 2$. If $q < 0$, put $\varepsilon := -q > 0$; every $y$ with $|y - q| < \varepsilon$ satisfies $y < q + \varepsilon = 0$ by [L2], hence $y \notin S$. If $q \ge 0$ and $q^2 > 2$, then $q \ne 0$ since $0^2 = 0 < 2$, so $q > 0$; put $\varepsilon := \min\{\, q,\ (q^2 - 2)(2q)^{-1} \,\} > 0$, again a definition by cases. Every $y$ with $|y - q| < \varepsilon$ satisfies $y > q - \varepsilon \ge 0$, so $y^2 > (q - \varepsilon)^2$ by [L4], and $(q-\varepsilon)^2 = q^2 - 2q\varepsilon + \varepsilon^2 \ge q^2 - 2q\varepsilon \ge q^2 - (q^2 - 2) = 2$, whence $y^2 > 2$ and $y \notin S$. In both cases a neighbourhood of $q$ misses $S$, so $\mathbb{Q} \setminus S$ is open in $\mathbb{Q}$.

For $r \in S$ put $B_r := \{\, y \in \mathbb{Q} : y < r \,\}$; each $B_r$ is open in $\mathbb{Q}$, since $y \in B_r$ and $\varepsilon := r - y > 0$ give, for every $z$ with $|z - y| < \varepsilon$, the inequality $z < y + \varepsilon = r$ by [L2].

The family $\mathcal{U} := \{\, B_r : r \in S \,\}$ is a cover of $S$ by sets open in $\mathbb{Q}$: given $q \in S$, step 1.2 supplies $r \in S$ with $q < r$, so $q \in B_r$.

$\mathcal{U}$ has no finite subfamily covering $S$: the empty subfamily fails because $S \ne \varnothing$ by step 1.1; and a nonempty finite subfamily is $\{B_{r_0}, \dots, B_{r_p}\}$ with every $r_i \in S$, so an induction on $p$ using the totality of the order of $\mathbb{Q}$ produces $R := \max\{r_0, \dots, r_p\}$, one of the $r_i$ and hence a member of $S$; for each $i$ one has $r_i \le R$, so $R < r_i$ fails and $R \notin B_{r_i}$. Thus the element $R$ of $S$ lies in no member of the subfamily.

The set $S$ is bounded by step 1.1 and closed in $\mathbb{Q}$ by step 1.3, and by steps 2.1 and 2.2 it is not compact in $\mathbb{Q}$, while $\mathbb{Q}$ is an ordered field by [L1]. So the claim [A1] fails at $F = \mathbb{Q}$ and is false.