Instead on relying on the more general results proved earlier we will reduce the statements to the corresponding statements for finite type $k$-algebras and cite results from the chapter on commutative algebra. \medskip\noindent Proof of (\ref{item-catenary}). This is local on $X$ by Topology, Lemma \ref{topology-lemma-catenary}. Thus we may assume $X = \Spec(A)$ where $A$ is a finite type $k$-algebra. We have to show that $A$ is catenary (Algebra, Lemma \ref{algebra-lemma-catenary}). We can reduce to $k[x_1, \ldots, x_n]$ using Algebra, Lemma \ref{algebra-lemma-quotient-catenary} and then apply Algebra, Lemma \ref{algebra-lemma-dimension-height-polynomial-ring}. Alternatively, this holds because $k$ is Cohen-Macaulay (trivially) and Cohen-Macaulay rings are universally catenary (Algebra, Lemma \ref{algebra-lemma-CM-ring-catenary}). \medskip\noindent Proof of (\ref{item-dimension-at-closed-point}). Choose an affine neighbourhood $U = \Spec(A)$ of $x$. Then $\dim_x(X) = \dim_x(U)$. Hence we reduce to the affine case, which is Algebra, Lemma \ref{algebra-lemma-dimension-closed-point-finite-type-field}. \medskip\noindent Proof of (\ref{item-dimension-irreducible}). It suffices to show that any two nonempty affine opens $U, U' \subset X$ have the same dimension (any finite chain of irreducible subsets meets an affine open). Pick a closed point $x$ of $X$ with $x \in U \cap U'$. This is possible because $X$ is irreducible, hence $U \cap U'$ is nonempty, hence there is such a closed point because $X$ is Jacobson by Lemma \ref{lemma-locally-finite-type-Jacobson}. Then $\dim(U) = \dim(\mathcal{O}_{X, x}) = \dim(U')$ by Algebra, Lemma \ref{algebra-lemma-dimension-spell-it-out} (strictly speaking you have to replace $X$ by its reduction before applying the lemma). \medskip\noindent Proof of (\ref{item-irreducible-maximal-chains}). Given a chain of irreducible closed subsets we can find an affine open $U \subset X$ which meets the smallest one. Thus the statement follows from Algebra, Lemma \ref{algebra-lemma-dimension-spell-it-out}  and $\dim(U) = \dim(X)$ which we have seen in (\ref{item-dimension-irreducible}). \medskip\noindent Proof of (\ref{item-dimension-irreducibles-passing-through}). Choose an affine neighbourhood $U = \Spec(A)$ of $x$. Then $\dim_x(X) = \dim_x(U)$. The rule $Z \mapsto Z \cap U$ is a bijection between irreducible components of $X$ passing through $x$ and irreducible components of $U$ passing through $x$. Also, $\dim(Z \cap U) = \dim(Z)$ for such $Z$ by (\ref{item-dimension-irreducible}). Hence the statement follows from Algebra, Lemma \ref{algebra-lemma-dimension-at-a-point-finite-type-over-field}. \medskip\noindent Proof of (\ref{item-dimension-irreducible-trdeg}). By (\ref{item-dimension-irreducible}) this reduces to the case where $X = \Spec(A)$ is affine. In this case it follows from Algebra, Lemma \ref{algebra-lemma-dimension-prime-polynomial-ring} applied to $A_{red}$. \medskip\noindent Proof of (\ref{item-immediate-specialization}). Let $Z = \overline{\{x\}} \supset Z' = \overline{\{x'\}}$. Then it follows from (\ref{item-irreducible-maximal-chains}) that $Z \supset Z'$ is the start of a maximal chain of irreducible closed subschemes in $Z$ and consequently $\dim(Z) = \dim(Z') + 1$. We conclude by (\ref{item-dimension-irreducible-trdeg}). \medskip\noindent Proof of (\ref{item-dimension-sup-trdeg}). A simple topological argument shows that $\dim(X) = \sup \dim(Z)$ where the supremum is over the irreducible components of $X$ (hint: use Topology, Lemma \ref{topology-lemma-irreducible}). Thus this follows from (\ref{item-dimension-irreducible-trdeg}). \medskip\noindent Proof of (\ref{item-specialization}). Part (a) follows from the fact that any open $U \subset X$ containing $x'$ also contains $x$. Part (b) follows because $\mathcal{O}_{X, x}$ is a localization of $\mathcal{O}_{X, x'}$ hence any chain of primes in $\mathcal{O}_{X, x}$ corresponds to a chain of primes in $\mathcal{O}_{X, x'}$ which can be extended by adding $\mathfrak m_{x'}$ at the end. Both (c) and (d) follow formally from (\ref{item-immediate-specialization}). \medskip\noindent Proof of (\ref{item-dimension-formula}). Choose an affine neighbourhood $U = \Spec(A)$ of $x$. Then $\dim_x(X) = \dim_x(U)$. Hence we reduce to the affine case, which is Algebra, Lemma \ref{algebra-lemma-dimension-at-a-point-finite-type-field}. \medskip\noindent Proof of (\ref{item-immediate-specialization-local-ring}). If $X$ is equidimensional (Topology, Definition \ref{topology-definition-equidimensional}) then $\dim(X)$ is equal to the dimension of every irreducible component of $X$, whence $\dim_x(X) = \dim(X) = \dim_{x'}(X)$ by (\ref{item-dimension-irreducibles-passing-through}). Thus this follows from (\ref{item-immediate-specialization}).