Is there a short proof for the Intermediate Value Theorem

The indermediate value theorem says:

Let $f:[a,b]\to \mathbb{R}$ be continuous and $f(a)<0$ and $f(b)>0$, then there exists a $\xi \in (a,b)$ such that $f(\xi)=0$.

You can prove it by using nested intervals:
You look at $f(\frac{a+b}{2})$, when it is bigger than null you look at $f$ on the interval $[a,\frac{a+b}{2}]$, if it is smaller than 0 we look at $[\frac{a+b}{2},b]$, when it is $0$ we are done. Lets denote the left endpoints with $a_n$ and the right endpoints with $b_n$.

As the diameter of our nested intervals is $(b-a)\cdot 2^{-n}$ which clearly converges to zero we have $$\lim_{n \to \infty} a_n =\lim_{n\to \infty} b_n=\xi$$ As $f$ is continuous we get $$\lim_{n\to \infty} f(a_n)=\lim_{n\to \infty} f(b_n)=f(\xi)$$ On the other hand we know $$f(a_n) < 0 \quad \forall n$$ and $$ f(b_n)>0 \quad \forall n $$ Hence we know $$\lim_{n\to \infty} f(a_n)\leq 0$$ and $$\lim_{n\to \infty} f(b_n)\geq 0$$ Hence $$0\leq f(\xi) \leq 0$$ Hence $f(\xi)=0$

You use that when $C_i$ is closed, bounded and non empty for all $i$ and $C_{i+1} \subset C_i$ for all $i$ then $$\bigcap_{i \in \mathbb{N}} C_i \neq \varnothing$$

The important part is to apply the definition of continuity:

$f(x)$ is continuous at point $c$ if $\forall \epsilon>0, \exists \delta>0 \quad S.T \quad |x-c|<\delta \Rightarrow |f(x)-f(c)|<0$ (Alternatively speaking, there always exist $\delta>0$ such that for any $\epsilon>0,|f(x)-f(c)|<0$).

Proof:

Define $c=sup\{x:f(x)\leq u\}$ ($\star$), and claim $f(c)=u$

Assum $x\in (-\delta+c,c+\delta),\delta>0$ ($x$ is in the $\epsilon$ neighborhood of $x$)

By definition of continuity $\forall \epsilon>0, |f(x)-f(c)|<\epsilon \Rightarrow -\epsilon+f(c)<f(x)<f(c)+\epsilon$

In the following, we will manipulate both side of this inequality and prove the intermediate value theorem by contradiction:

1. If $f(c)>u$, then $f(c)-u>0$, so set $\epsilon=f(c)-u \Rightarrow f(x)>f(c)-\epsilon=f(c)-(f(c)-u)=u$(use the left side of the inequality ) $\Rightarrow \forall x\in (-\delta+c,c+\delta),f(x)>u$, so it means that $(c-\delta)$ is the least upper bound of the set $\{x:f(x)\leq u\}$, which contradicts with the definition of $c$(also a least upper bound and it's not possible to have 2 least upper bounds at the same time.)
2. If $f(c)<u$, then $u-f(c)>0$,so set $\epsilon=u-f(c) \Rightarrow f(x)<f(c)+\epsilon=f(c)+(u-f(c))=u$(use the right side of the inequality)$\Rightarrow \forall x\in (-\delta+c,c+\delta),f(x)<u$. This means that there exist $x>c$ such that $f(x)<u$, which contradicts with the definition of $c$ again. (because $c$ is the sup of the set $\{x:f(x)\leq u\}$)

If you set $u=0$, then it's the answer of your question. Hope this can help.

This theorem is a case when the most intuitive proof requires relatively advanced technique. here's a proof using general topology.

Consider a continuous function $f: \mathbb{R} \to \mathbb{R}$ which takes values in $a, b$ but not $c \in (a, b)$. Then $f$ factors through the embedding $i: \mathbb{R} \setminus \{c\} \to \mathbb{R}$: $f = i \circ \hat{f}$, where $\hat{f}: \mathbb{R} \to \mathbb{R} \setminus \{c\}$. $\hat{f}$ is continuous: any open set of $\mathbb{R} \setminus \{c\}$ is obtained by removing $c$ from an open subset of the whole $\mathbb{R}$. But since $f$ is nowhere equal $c$, removing this point doesn't affect the preimage, which is open. $\mathbb{R}$ is path-connected (and thus connected), but $\mathbb{R} \setminus \{c\}$ has two connected components. Therefore, the image of $\hat{f}$ must lie to the one side of $c$ - contradiction.