Small Inductive Dimension is a topological invariant I have started to read Engelking's book "Theory of Dimesions: Finite and Infinite" and the author states (pp. $3$) that:

The small inductive dimension (also called Menger-Urysohn dimension) is a topological invariant, i.e. $h\colon X\to Y$ homeo implies $\mathrm{ind}X=\mathrm{ind}Y$.

Here I post my tentative proof because I want to be sure it is correct.
Any suggestion is appreciated.
As I understand it, the proof relies on the fact that any homeomorphism preserves boundaries.
Notation: Here $U(x)$, $V(x)$ denote an open nhbd of point $x\in X$; $Fr[.]$ is the usual boundary operator.
For the convenience of the reader, this is the definition given in the text:

Definition:
(MU$1$) $\mathrm{ind} X=-1$ iff $X=\emptyset$.
(MU$2$) $\mathrm{ind} X\le n$ iff $\forall x\in X$ $\forall U(x)$ $\exists V(x)$ s.t. $V(x)\subseteq U(x)$ and $\mathrm{ind} Fr[V(x)]\le n-1$.
(MU$3$) $\mathrm{ind}X=n$ iff $n-1<\mathrm{ind}X\le n$.
(MU$4$) $\mathrm{ind} X=\infty$ iff $\mathrm{ind}X>n $ for every $n=-1,0,\dots$

Attempt:
(MU$1$) It is obvious.
(MU$2$) Induction on $n$. The case $n=0$ follows by the fact that $Fr[V(x)]=\emptyset$; if $n\ge 1$, we want to prove it for $n+1$. Assume $\mathrm{ind}X\le n+1$, then $\mathrm{ind}Fr[V(x)]\le n$ and by the inductive hypothesis we get
\begin{equation}\tag{*}
\label{eqn:*}
\mathrm{ind}Fr[V(x)]=\mathrm{ind}\,h(Fr[V(x)])=\mathrm{ind}Fr(h[V(x)])
\end{equation}
and we are done.
The remaining cases (MU$3$) and (MU$4$) are obtained in a similar way using \eqref{eqn:*}.
Does it work?
Thank you in advance for your help.
 A: If $h:X \to Y$ is a homeomorphism then $\operatorname{ind}(X) = \operatorname{ind}(Y)$. 
Maybe first show: $$\phi_n:\text{for all homeomorphisms } h \text { between all spaces } X \text{ and } Y: \operatorname{ind}(X) \le n \implies \operatorname{ind}(Y) \le n$$ by induction on $n \in \{-1\}\cup \mathbb{N}$.
If $n=-1$, we have $\operatorname{ind}(X)\le -1 \iff X =\emptyset$ and as $h$ is a homeomorphism (so a bijection that preserves cardinality), $Y=\emptyset$ and so also $\operatorname{ind}(Y) =-1 \le -1$. This covers the base case.
If $\phi_n$ holds, then so does $\phi_{n+1}$: Assume $\operatorname{ind}(X) \le n+1$ and let $y \in Y$, and $U_y$ be an open neighbourhood of $y$. As $h$ is a homeomorphism, there is some $x \in X$ and an open $U_x$ containing $x$, such that $h(x)=y$ and $h[U_x]=U_y$.
Because of the assumption $\operatorname{ind}(X) \le n+1$ there exists (by clause MU2 of the definition) an open neighbourhood $V_x$ of $x$ such that $V_x \subseteq U_x$ and $\operatorname{ind}(\operatorname{Fr}(V_x)) \le n$. The restriction of $h$ to $\operatorname{Fr}(V_x)$ is a homeomorphism with this subspace to $\operatorname{Fr}(h[V_x])$, by standard topological facts on homeomorphisms. So by the induction assumption $\phi_n$ we know that 
$$\operatorname{ind}(\operatorname{Fr}(h[V_x]) \le n$$
and this shows that $h[V_x]$ is the required open neighbourhood that allows us, again from MU2, to conclude that $\operatorname{ind}(Y) \le n$, as $y$ and $U_y$ were arbitrary. 
So $\phi_n$ holds by induction.
So if $X$ and $Y$ are homeomorphic spaces via $h$, $Y$ and $X$ are homeomorphic via $h^{-1}$ so for all $n \in \{-1\} \cup \mathbb{N}$ we have $\operatorname{ind}(X)=\operatorname{ind}(Y)$ from both inequalities. The infinite case is an immediate corollary: if $X$ is infinite dimensional, we couldn't have $\operatorname{Y} \le n$ for some $n$ as $\phi_n$ applied to $h^{-1}$ gives a direct contradiction.
