The multiplicity of the Laplacian eigenvalue $0$ equals the number of connected components

For a vector $x=(x_v{)}_{v\in V(G)}$, the Laplacian quadratic form is $x^{\mathsf{T}}Lx={\sum }_{\{u,v\}\in E(G)}(x_u-x_v{)}^2$, because expanding $x^{\mathsf{T}}(D-A)x$ groups the diagonal terms by degrees and subtracts each edge contribution twice. Therefore, if $Lx=0$, then [L2] gives $0=x^{\mathsf{T}}Lx$, so every summand vanishes and $x_u=x_v$ on every edge. Conversely, if $x$ is constant on every edge, then for each vertex $u$ every neighbor $v$ of $u$ satisfies $x_v=x_u$, so [F2] gives $(Lx{)}_u=\mathrm{deg}(u)x_u-{\sum }_{v\sim u}{x}_v=0$. Hence $Lx=0$.

By [F1], a function on $V(G)$ is constant on every edge exactly when it is constant on each connected component. By [L1], the indicator vectors ${\mathbf{1}}_{C_1},\dots ,{\mathbf{1}}_{C_r}$ are linearly independent and every vector constant on components is a linear combination of them. So step 1.1 shows that these indicator vectors form a basis of $\ker L$, and hence $\dim \ker L=r$.

The final equivalence with multiplicity of the eigenvalue $0$ follows from [L3], because $L$ is real symmetric.