Does the given $h$ exhibit the homeomorphism between $\mathbb{R}^{\omega}$ and itself with box topologies?

Does the given $h$ exhibit the homeomorphism between $\mathbb{R}^{\omega}$
and itself with box topologies?

(Munkres, p. 118, Problem 8)
Give $h: \mathbb{R}^{\omega} \to \mathbb{R}^{\omega}$ $$ h(x_1,x_2,\dots)
= h(ax_1 +b,ax_2+b,\dots) $$ where $a,b\in \mathbb{R}$.
Does the given $h$ exhibit the homeomorphism between $\mathbb{R}^{\omega}$
and itself with box topologies?
My proof goes like this:
Since each $x_i \mapsto ax_i +b$ is continuous, for each open ball of
$B_i(ax_i +b, \epsilon_i)$, we can find $B_i(x,\delta_i)$ such that
$aB_i(x,\delta_i) +b \subseteq B_i(ax_i +b, \epsilon_i)$. Now construct $$
B_1(x_i,\delta_i) \times B_2(x_i,\delta_i) \times \cdots. $$ Then $$
h(B_1(x_i,\delta_i) \times B_2(x_i,\delta_i) \times \cdots) \subseteq
B_1(ax_1 +b, \epsilon_1) \times B_2(ax_2 +b, \epsilon_2) \times \cdots $$
proving the continuity.
Would this be valid?