My teacher in course in Mat-2.3140 of Aalto University claims that 'All polyhedrons are convex sets' here. This premise was in a false-or-not-problem 'The feasible set of linear integer problem is polyhedron'. You can see below the screenshot of the solution.
Wikipedia shows nonconvex polyhedrons such as orthogonal polyhedron here.
What should I now believe? Is polyhedron convex or not?
Definitions on the lecture slides (p.8, L4)
Polyhedron is such that $$P=\{\bar x\in \mathbb R^n | A \bar x\geq \bar b\}, A\in \mathbb R^{m\times n},\bar b\in \mathbb R^m$$ and a convex function $f(x)$ must satisfy $$f(\lambda \bar x+(1-\lambda)\bar y)\leq \lambda f(\bar x)+(1-\lambda) f(\bar y) \text{, } \forall \bar x, \bar y, \lambda \in [0,1]$$ and a convex set $C$ is such that $$\bar x, \bar y \in C\rightarrow \lambda \bar x+(1-\lambda)\bar y\in C.$$
I suspect you are confused with the definition. Usually a a polyhedron is defined by specifying a finite subset of $n-1$ dimensional affine subspaces in $\mathbb{R}^{n}$. In this way what you get is always convex. This is the definition people use when work on combinatorical topology or algebraic combinatorics. You should confirm this with your teacher though.