The main problem is to prove the following theorem.
Let $X$ be a $\sigma$-compact Hausdorff space. If every compact subspace of $X$ has topological dimension at most $m$, then so does $X$.
Also, we have a definition.
$X$ is said to be $\sigma$-compact if there is a countable collection of compact subspaces of $X$ whose interiors cover $X$.
Let $\mathbb{A}$ be an open cover of $X$. The strategy is to find an open cover $\mathbb{B}$ of $X$ refining $\mathbb{A}$ that has order at most $m+1$.
The following is part (c), which I can't understand.
Suppose $n \geq 0$ and $\mathbb{B}_n$ is an open covering of $X$ refining $\mathbb{B}_0$ such that $\mathbb{B}_n$ has order at most $m+1$ at points of $X_n$. Choose an open covering $\mathbb{C}$ of $X$ refining $\mathbb{B}_n$ that has order at most $m + 1$ at points of $X_{n+1}$. Choose $f : \mathbb{C} \to \mathbb{B}_n$ so that $C \subset f(C)$. For $B \in \mathbb{B}_n$, let $D(B)$ be the union of those $C$ for which $f(C) = B$. Let $\mathbb{B}_{n+1}$ consist of all sets $B \in \mathbb{B}_n$ for which $B \cap X_{n-1} \neq \emptyset$, along with all sets $D(B)$ for which $B \in \mathbb{B}_n$ and $B \cap X_{n-1} = \emptyset$. Show that $\mathbb{B}_{n+1}$ is an open covering of $X$ that refines $\mathbb{B}_n$ and has order at most $m+1$ at points of $X_{n+1}$.
And the following are part (a), (b), (d) of the same question, respectively.
(a). Show that $X = \bigcup X_n$, where $X_n$ is compact and $X_n \subset$ Int $X_{n+1}$ for each $n$. Let $X_0 = \emptyset$.
(b). Find an open covering $\mathbb{B}_0$ of $X$ refining $\mathbb{A}$ such that for each $n$, each element of $\mathbb{B}_0$ that intersects $X_n$ lies in $X_{n+1}$.
(d). Define $\mathbb{B}$ as follows: Given a set $B$, it belongs to $\mathbb{B}$ if there is an $N$ such that $B \in \mathbb{B}_n$ for all $n \geq N$.
(a), (b) is done, I solved them. But I don't understand what (c) is trying to do. It seems that I should inductively define $\mathbb{B}_n$ such that it refines $\mathbb{B}_0$ and has order at most $m+1$ at points of $X_n$. So at the beginning of (c), $\mathbb{B}_n$ satisfying the condition should be given and I should use it to make a $\mathbb{B}_{n+1}$ that refines $\mathbb{B}_n$ and has order at most $m+1$ at points of $X_{n+1}$. But at the second sentence, the question already assumes that a collection $\mathbb{C}$ satisfying this condition exists('Choose an open covering $\mathbb{C}$ of $X$ refining $\mathbb{B}_n$ that has order at most $m + 1$ at points of $X_{n+1}$'). It seems to be a nonsense. And I have no idea how to prove that the collection $\mathbb{C}$ satisfying the condition does exist.
Let us start with
Lemma. Let $A \subset X$ be closed. Then $\dim A \le m$ if and only if for each open covering $\mathcal U$ of $X$ there exists an open covering $\mathcal V$ of $X$ which refines $\mathcal U$ and has order at most $m+1$ at all points of $A$. [The latter was not explicitly defined by Munkres, but it means that some point of $A$ lies in $m+1$ elements of $\mathcal V$ and no point of $A$ lies in more than $m+1$ elements of $\mathcal V$.]
"If part" : Let $\mathcal U'$ be an open covering of $A$. For each $U' \in \mathcal U'$ choose an open $U \subset X$ such that $U \cap A = U'$. These $U$ together with $X \setminus A$ form an open covering $\mathcal U$ of $X$. There exists an open covering $\mathcal V$ of $X$ which refines $\mathcal U$ and has order at most $m+1$ at all points of $A$. Let $\mathcal V'$ be the set of all nonempty $V \cap A$ with $V \in \mathcal V$. This is an open covering of $A$. Since each $V$ admits $U \in \mathcal U$ such that $V \subset U$, the same is true for the intersections: $V \cap A \subset U \cap A$. Since $V \cap A \ne \emptyset$, we cannot have $V \subset X \setminus A$, hence we must have $U \cap A \in \mathcal U'$. Thus $\mathcal V'$ refines $\mathcal U'$. But by construction $\mathcal V'$ has order at most $m+1$.
"Only if part" : Let $\mathcal U$ be an open covering of $X$. Then the set $\mathcal U'$ of all nonempty $U \cap A$ with $U \in \mathcal U$ is an open covering of $A$. There exists an open covering $\mathcal V'$ of $A$ which refines $\mathcal U'$ and has order at most $m+1$. For each $V' \in \mathcal V'$ there exists $U_{V'} \in \mathcal U$ such that $V' \subset U_{V'} \cap A$. Let $V \subset X$ be open such that $V \cap A = V'$ and define $W_{V'} = U_{V'} \cap V$. The set of these $W_{V'}$ together with the set of all nonempty $U \setminus A$ with $U \in \mathcal U$ forms an open covering of $X$ which refines $\mathcal U$. By construction it has order at most $m+1$ at all points of $A$.
In (c) you start with an open covering $\mathbb{B}_n$ of $X$ which refines $\mathbb{B}_0$ and has order at most $m+1$ at points of $X_n$. Since $X_{n+1}$ is compact and thus has $\dim X_{n+1} \le m$, the lemma gives us an open covering $\mathbb{C}$ of $X$ refining $\mathbb{B}_n$ that has order at most $m + 1$ at points of $X_{n+1}$.