Let I be a real interval. Let a,b in I such that (a..b) is an open interval. Let f:I->R be a real function which is continuous on (a..b). Let k in R lie between f(a) and f(b). That is, either f(a)<k<f(b) or f(b)<k<f(a). Then EEc in (a..b) such that f(c)=k.

Found it:
Let $I$ be a real interval. Let $a,b\in I$ such that $\left(a..b\right)$ is an open interval. Let $f:I\to \mathbb{R}$ be a real function which is continuous on $\left(a..b\right)$. Let $k\in \mathbb{R}$ lie between $f\left(a\right)$ and $f\left(b\right)$. That is, either $f\left(a\right) or $f\left(b\right). Then $\mathrm{\exists }c\in \left(a..b\right)$ such that $f\left(c\right)=k$.

Shouldn't this suppose continuity on the closed interval $\left[a,b\right]$? Otherwise consider $f:\left[-1,1\right]\to \mathbb{R}$ such that
$f\left(x\right)=\left\{\begin{array}{ll}\mathrm{arcsin}\left(x\right)& \text{-1
where $a=-1$ and $b=1$.
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plomet6a
Yes, I agree. The given statement is incorrect and your counterexample shows it.

An even simpler example might just be
$f\left(x\right)=\left\{\begin{array}{ll}0,& x=-1\\ 2,& -1
with $k=\frac{1}{2}$.