I believe this is an alternative way to prove intermediate value theorem. Do not know if it is harder.
Let $D=[a,b]$ be a closed interval, where $a<b$ are real numbers. Let $f \colon D \rightarrow \mathbb{R}$ be a continuous function such that $f(a)<f(b)$. Consider $d \in \mathbb{R}$ such that $f(a)<d<f(b)$ and consider the set $S=\{x \in [a,b] \colon f(x) \leq d\}$.
We know that $S$ is not empty since $f(a) < d \implies a \in S$. We also know $S$ is bounded since it is a subset of $D$ ($S\subseteq D$). Thus by Axiom of Completeness, $\sup(S)$ exists.
How can I prove $c=\sup(S)$ satisfies $f(c)=d$? (without IVT obviously)
Can we have $f(c)<d$? If that happens, then $c<b$ (because $f(b)\geqslant d$). Take $\delta>0$ such that$$|x-c|<\delta\wedge x\in[a,b]\implies\bigl|f(x)-f(c)\bigr|<d-f(c).$$Then, if $x_0\in(c,c+\delta)\cap[a,b]$ (this set is not empty, since $c<b$), you have\begin{align}f(x_0)&=f(x_0)-f(c)+f(c)\\&\leqslant\bigl|f(x_0)-f(c)\bigr|+f(c)\\&<d-f(c)+f(c)\\&=d.\end{align}But then $x_0\in S$ and $x_0>c$, which is impossible, since $c=\sup S$.
What about having $f(c)>d$? Then, take $\delta>0$ such that$$|x-c|<\delta\wedge x\in[a,b]\implies\bigl|f(x)-f(c)\bigr|<f(c)-d.$$Then, if $x_0\in(c-\delta,c)\cap[a,b]$,\begin{align}f(x_0)&=f(x_0)-f(c)+f(c)\\&\geqslant-\bigl|f(x_0)-f(c)\bigr|+f(c)\\&>-\bigl(f(c)-d\bigr)+f(c)\\&=d.\end{align}So, there is no element of $S$ in $(c-\delta,c)\cap[a,b]$, which is impossible, again because $c=\sup S$.
So, since we don't have $f(c)\ne d$, we must have $f(c)=d$.