Why is $\operatorname{colim} F \cong \pi_0\left (\int F\right )$?

I think I finally saw the key observation needed in this. For any diagram $F: \mathsf{C\to Set}$ and cone $\lambda: F \Longrightarrow X$ for $X \in \mathsf{Set}$ we have individual morphisms $\lambda_c: Fc \to X$ indexed by the objects of $\mathsf{C}$ and the compatibility condition that states for any $f \in \mathsf{C}(c,d)$ we have that $\lambda_c = \lambda_d\circ Ff$.

The key observation comes in seeing each leg of the cone $\lambda_c: Fc \to X$ as really a slice of the mapping $\lambda: \int F \to X$. In other words, an element $x \in Fc$ can really be thought of as the pair $(c,x) \in \int F$ and $\lambda_c(x) = \lambda(c,x)$. Most importantly the compatibility condition shows that any two elements $(c,x)$ and $(d,y)$ in $\int F$ get mapped to the same element of $X$ as long as there's a morphism $f \in \mathsf{C}(c,d)$ where $Ff(x) = y$. By extension, any two elements of $\int F$ joined by a finite sequence of such morphisms gets mapped to the same element of $X$. Thus we can conclude $\lambda:\int F \to X$ is constant on the path components of $\int F$ and therefore descends to a mapping $\tilde{\lambda}: \pi_0\left (\int F\right ) \to X$. This mapping is unique since a different mapping would not be compatible with the cone $\lambda$ specified above; i.e. this is the only such mapping that $\lambda$ can descend to.

Furthermore, this induces a cone $\eta:F \Longrightarrow \pi_0\left (\int F\right )$ which assigns $x \in Fc$ its path component $[(c,x)]$. We see that any cone $\lambda:F\Longrightarrow X$ factors uniquely as $\lambda = \tilde{\lambda}\circ \eta$.