If $E$ is a contractible space on which the (Edit: topological) group $G$ acts freely, then $E/G$ is a classifying space for $G$.
A better, but still false, version:
If $E$ is a free, contractible $G$-space and the quotient map $E\to E/G$ admits local slices, then $E/G$ is a classifying space for $G$.
(Here "admits local slices" means that there's a covering of $E/G$ by open sets $U_i$ such that there exist continuous sections $U_i \to E$ of the quotient map.)
The simplest counterexample is: let $G^i$ denote $G$ with the indiscrete topology (Edit: and assume $G$ itself is not indiscrete). Then G acts on $G^i$ by translation and $G^i$ is contractible (for the same reason: any map into an indiscrete space is continuous). Since $G^i/G$ is a point, there's a (global) section, but it cannot be a classifying space for $G$ (unless $G=\{1\}$). The way to correct things is to require that the translation map $E\times_{E/G} E \to G$, sending a pair $(e_1, e_2)$ to the unique $g\in G$ satisfying $ge_1 = e_2$, is actually continuous.
Of course the heart of the matter here is the corresponding false belief(s) regarding when the quotient map by a group action is a principal bundle.