Suppose $D_i$ ~ $Bern(q)$. Let $\hat{q} = \frac{1}{N}\Sigma{D_i}$. Find the distribution of $\frac{N(\hat{q} - q)^2}{1-q}$ as N goes to infinity.
By Central Limit Theorem I have $dlim(\sqrt{N}(\hat{q} - q)) = N(0, q(1-q))$. Apply Continuous Mapping Theorem, $N(\hat{q} - q)^2$ converges to $[N(0, q(1-q)]^2$. Is this Chi-squared distribution? If I divide by $q(1-q)$, what distribution would I get?
As you already stated, by using the CLT, continuous mapping and Slutsky, you have that $$ \frac{ \sqrt{n}( \hat{q} - q) }{\sqrt{q(1-q)} }\xrightarrow{D} Z, $$ such that $\sim \mathcal{N}(0,1)$. Hence, by using the continuous mapping for $g(x) = x^2$ once again, you have that $$ g \left( \frac{ \sqrt{n}( \hat{q} - q) }{\sqrt{q(1-q)} } \right) \xrightarrow{D} g(Z)=Z^2. $$ One of the definitions of the $\chi^2$ distribution is a square of the standard normal random variable, thus $$ g(Z) =Z^2\sim\chi_{(1)}^2. $$