We shall prove that the set $S = \{x : Ax \leq b \}$ is convex. The same proof works for $Ax \geq b$.
Consider any $n \in \mathbb{N}$. Consider $n$ elements in $S$. Call them $x_1,x_2,\ldots,x_n$ i.e. $x_k \in S$ for $k \in \{1,2,\ldots,n\}$.
Consider $n$ scalars $\lambda_j$ such that $\displaystyle \sum_{j=1}^{n} \lambda_j = 1$. We want to show that $x = \displaystyle \sum_{j=1}^{n} \lambda_j x_j \in S$.
Since $x_j \in S$, we have $Ax_j \leq b$. Hence, we get $Ax = A \left( \displaystyle \sum_{j=1}^{n} \lambda_j x_j \right) = \displaystyle \sum_{j=1}^{n} A \left(\lambda_j x_j \right) = \displaystyle \sum_{j=1}^{n} \lambda_j A x_j \leq \displaystyle \sum_{j=1}^{n}\lambda_j b = \left(\displaystyle \sum_{j=1}^{n} \lambda_j \right) b = b$ Hence, we have $Ax \leq b.$ Hence, $S$ is a convex set.