Below is my little proof of lub axiom
DEFINITIONS :
$A$ be any non empty subset of $\mathbb{R}$ bounded above
$B$ be the set of all real upper bounds of $A$
$x \in B\ \Leftrightarrow \forall\ a \in A, a \leq x$
$B^{*}$ be the set of all reals which are not real upper bounds of $A$
$x \in B^{*}\ \Leftrightarrow \exists \ a \in A, a > x$
LEMMA 1 : $x \in B^{*} \implies \bigg[ \forall x' < x, x' \in B^{*} \bigg]$
Proof : (easy)
LEMMA 2 : $x \in B \implies \bigg[ \forall x' > x, x' \in B \bigg]$
Proof : (easy)
LEMMA 3 : $\exists\ \xi \in \mathbb{R}, \bigg[ (\forall\ x > \xi, x \in B) ⊻ (\forall\ x < \xi, x \in B^{*}) \bigg]$
Proof :
Choose any $x_1 \in B^{*}$ and any $x_2 \in B$
Consider closed interval $[x_1,x_2]$
Obviously, $\forall\ x \in [x_1,x_2], \bigg[\ (x \in B^{*}) ⊻ (x \in B)\ \bigg]$
Create a function $f:[x_1,x_2] \to \{0,1\}$ giving $0$ output to $x \in B^{*}$ and $1$ output to $x \in B$
Obviously, $x_1$ has output $0$ and $x_2$ has output $1$
By $\Huge\text{Theorem @}$ on continuous functions, there is discontinuity in $f:[x_1,x_2] \to \{0,1\}$
By (LEMMA 1) and (LEMMA 2),
There is one and only one discontinuity in $f:[x_1,x_2] \to \{0,1\}$, say at $\xi \in [x_1,x_2]$
s̲i̲n̲c̲e̲ ̲m̲o̲r̲e̲ ̲t̲h̲a̲n̲ ̲o̲n̲e̲ ̲d̲i̲s̲c̲o̲n̲t̲i̲n̲u̲i̲t̲y̲ ̲v̲i̲o̲l̲a̲t̲e̲s̲ ̲(̲L̲E̲M̲M̲A̲ ̲1̲)̲ ̲a̲n̲d̲ ̲(̲L̲E̲M̲M̲A̲ ̲2̲)̲
$$\text{}$$ $$\text{AND}$$ $$\text{in closed interval $[x_1,x_2]$; the following three statements are true :}$$ $$\text{}$$
$\forall\ x < \xi, x \in B^{*} \tag1$
s̲i̲n̲c̲e̲ $x < \xi$ ̲m̲u̲s̲t̲ ̲h̲a̲v̲e̲ ̲o̲u̲t̲p̲u̲t̲ $0$, e̲l̲s̲e̲ ̲(̲L̲E̲M̲M̲A̲ ̲1̲)̲ ̲a̲n̲d̲ ̲(̲L̲E̲M̲M̲A̲ ̲2̲)̲ ̲g̲e̲t̲s̲ ̲v̲i̲o̲l̲a̲t̲e̲d̲
$\forall\ x > \xi, x \in B \tag2$
s̲i̲n̲c̲e̲ ̲$x > \xi$ m̲u̲s̲t̲ ̲h̲a̲v̲e̲ ̲o̲u̲t̲p̲u̲t̲ $1$, e̲l̲s̲e̲ ̲(̲L̲E̲M̲M̲A̲ ̲1̲)̲ ̲a̲n̲d̲ ̲(̲L̲E̲M̲M̲A̲ ̲2̲)̲ ̲g̲e̲t̲s̲ ̲v̲i̲o̲l̲a̲t̲e̲d̲
$(\xi \in B^{*}) ⊻ (\xi \in B) \tag3$
s̲i̲n̲c̲e̲ $\xi$ ̲h̲a̲s̲ ̲o̲u̲t̲p̲u̲t̲ ̲e̲i̲t̲h̲e̲r̲ $0$ ̲o̲r̲ $1$
Thus:
$\exists\ \xi \in \mathbb{R}, \bigg[ (\forall\ x > \xi, x \in B) ⊻ (\forall\ x < \xi, x \in B^{*}) \bigg]$
LEMMA 4 : $\xi \in B$
Proof : (easy)
Theorem (lub axiom) : $\xi$ is lub $A$
$\Huge\text{Questions}$
$(1)$ Does Theorem @ use lub axiom in any way ?
$(2)$ In my entire proof, am I using lub axiom in any way ?
$(3)$ Is the proof perfectly correct ?
The "Theorem" must use the LUB axiom because the theorem is false on the rationals. Define a function $ f $ on the rationals with $ f(r) = 0 $ if $ r^2 < 2 $ and $f(r)=1$ otherwise. $ f $ is continuous as a function on $ \mathbb{Q} $.