Bayesian Statistics - Proof that $\hat{\theta}(y) = \arg \max_{\theta} f_{\Theta|Y}(\theta|y)$ for the continuous 0-1 loss function

Currently taking statistics class as part of my Engineering degree, and right now we covered Bayesian Statistics. One Theorem that had come up is: $$\text{Theorem}: \text{for the continuous case of the 0-1 loss function: } \hat{\theta} = \arg \max_\theta \{f_{\Theta|Y}(\theta|y)\}$$ The professor's Proof was to use the MAP rule and just get: $$ \hat{\theta} = \arg \max_\theta \{f_{\Theta|Y}(\theta|y)\} $$ The problem is, he never said what the "MAP rule" is, and I can't find a good explanation anywhere. Using: $$ c(E) = \cases{1 \;\; |E| \ge \varepsilon \\0 \;\; |E| 0$$ I tried solving it myself, and as shown at Under the 0-1 loss function, the Bayesian estimator is the mode of the posterior distribution , I got the same result: $$ 1 - f_{\Theta|Y}(\hat{\theta}|y) $$ But, this is only true for $\varepsilon \longrightarrow 0$ . Finally, my question is, is there an easy/intuitive way to prove (or at least explain intuitively) what my professor did? Adding the Proof from class: