Example of locally contractible topological space where Čech cohomology does not coincide with singular cohomology I believe that it is shown in EH Spanier's "Algebraic Topology" that if  is paracompact and locally contractible, then singular cohomology and Čech cohomology of  coincide, with coefficients in any abelian group.
I can see why being locally contractible helps but can't think of why you would need paracompact as well.
I would love an example of a space where these cohomology theories fail to coincide due to the space not being paracompact.
 A: I don't have an example for you. I imagine such a thing may be pretty strange. However, as I understand it, paracompactness is used to ensure the homomorphism between singular and Cech cohomology represents sufficiently fine open covers of the space. This is required to make the "standard" proof work.
When a space $X$ is paracompact Hausdorff every open cover $\mathscr{U}$ of $X$ is normal in the sense that there is a locally finite partition of unity $\{\phi_{U}\}_{U\in \mathscr{U}}$ subordinate to $\mathscr{U}$. In Cech/shape theory, this is precisely what is need to construct "canonical maps" $p_{\mathscr{U}}:X\to |N(\mathscr{U})|$ to the nerve (the defining property of a canonical map is that $p_{\mathscr{U}}^{-1}(st(U,|N(\mathscr{U})|))\subseteq U$ for each $U\in\mathscr{U}$). Explicitly, you can define a canonical map from $\{\phi_{U}\}_{U\in \mathscr{U}}$ as follows: for $U\in \mathscr{U}$ and $x\in U$, determine $p_{\mathscr{U}}(x)$ by requiring its barycentric coordinate belonging to the vertex $U$ of $|N(\mathscr{U})|$ to be $\phi_{U}(x)$. Conversely, a canonical map $p_{\mathscr{U}}:X\to |N(\mathscr{U})|$ determines a partition of unity subordinate to $\mathscr{U}$. So you can't really have one without the other. More details can be found in the appendix of Mardesic and Segal's Shape Theory book.
The homomorphism $H^n(|N(\mathscr{U})|)\to H^n(X)$ is induced by any choice of canonical map $p_{\mathscr{U}}$. These induce the natural homomorphism $\phi:\check{H}^n(X)\to H^n(X)$ on the direct limit. Of course, you can still construct $\phi$ if $X$ is not paracompact by restricting to normal open covers and it very well might still be an isomorphism. However, the "standard" proofs used to show homomorphisms like $\phi$ are isomorphisms usually involve star-refining a given cover several times so that the refinements are covers consisting of contractible sets or satisfying some other connectedness condition. If $X$ is locally contractible, you can always refine to a cover of contractible sets but without paracompactness you may not be able to take a star-refinement. Moreover, if $X$ is not paracompact, you have no way of knowing if your arbitrarily fine covers consisting of locally contractible sets tell you anything about $\phi$ because these covers may not have normal refinements.
