Showing $f: \prod_{i \ge 1} \{0,2\}_i \rightarrow C$ is an Open Mapping into the Cantor Set Setting: Let $C$ denote the Cantor Set and $X = \prod_{i \ge 1} \{0,2\}_i$.  Let $X$ be given the product topology.  Consider $f : X \rightarrow C$ s.t. if $p = \left\langle x_1, x_2 , \ldots \right\rangle  \in X$, then $f(p) = \sum_{n \ge 1} x_n / 3^n  \in  C$.  My ultimate goal is to show that $f$ is a homeomorphism.  I've already shown that $f$ is continuous and bijective.  It remains to be shown that $f^{-1}$  is continuous.
Goal:  Show that $f^{-1}$ is continuous (or equivalently, show that $f$ is an open mapping).
Attempt:


*

*Let $W$ be non-empty and open in $X$ under the product topology.

*Then $f^{-1}$ is continuous if $(f^{-1})^{-1}(W) = f(W)$ is open in $C$.

*Since $W$ is open in $X$, we have that $W = \prod_{i \ge 1} W_i$ s.t. (i) $W_i \ne X \implies W_i = U \cap \{0,2\}$ s.t. $U$ is open in $\mathbb{R}$ and (ii) all but a finite number of the $W_i$ are equal to $\{0,2\}$.

*Then the last step yields that any index of $W$ could be $\{0\}$, $\{2\}$, or $\{0,2\}$ (with the former two options only occurring at most a finite number of times).

*Then $f(W) = \{\sum_{n \ge 1} a_n / 3^n :$ a finite number of the $a_n$ are fixed as either $0$ or $2\}$.
But then why must $f(W)$ be open in $C$?
 A: The goal is reached quicker by showing that $f$ is a closed mapping:


*

*$X$ is quasicompact, hence every closed subset $F$ of $X$ is quasicompact.

*$f$ is continuous, hence $f(F)$ is quasicompact.

*$C$ is Hausdorff, hence every quasicompact subset of $C$ is closed.


Thus $f$ maps closed subsets of $X$ to closed subsets of $C$, and that means $f^{-1}$ is continuous.
Along the chosen path, showing that $f$ is open, we reach the goal with a little more effort.
Note that your characterisation of open sets in $X$ isn't accurate, every open subset of $X$ is a union of sets of the specified structure. But as any union of open sets is again open, it suffices to show that the image of such sets is open.
More, it suffices to show that $f(W)$ is open for every $W$ of the form
$$W = \bigcap_{i=1}^n \pi_i^{-1}(\{a_i\}),\tag{1}$$
where $a_1,\dotsc,a_n \in \{0,2\}$ are arbitrarily chosen, since these sets form a basis of the topology on $X$.
Suppose first that in $(1)$, there are $j,k$ with $a_j = 0$ and $a_k = 2$. Let $m_0 = \max \{ i : 1 \leqslant i \leqslant n, a_i = 0\}$ and $m_2 = \max \{ i : 1 \leqslant i \leqslant n, a_i = 2\}$. Let $b_i = a_i$ for $i < m_2$, $b_{m_2} = 0$, and $b_i = 2$ for $m_2 < i \leqslant n$, and $c_i = a_i$ for $i < m_0$, $c_{m_0} = 2$, and $c_i = 0$ for $m_0 < i \leqslant n$.
Let $h(a) = \sum_{i=1}^n a_i\cdot 3^{-i}$, and analogously define $h(b)$ and $h(c)$. If $x\notin W$, there is a smallest $j \in \{1,\dotsc,n\}$ with $x_j \neq a_j$. If $x_j < a_j$, then
$$\sum_{i=1}^n x_i \cdot 3^{-i} \leqslant h(b),$$
and consequently
$$f(x) = \sum_{i=1}^\infty x_i\cdot 3^{-i} \leqslant h(b) + \sum_{i=n+1}^\infty 2\cdot 3^{-i} = h(b) + \frac{2}{3^{n+1}}\cdot \frac{1}{1-\frac{1}{3}} = h(b) + 3^{-n}.$$
If $x_j > a_j$, then
$$\sum_{i=1}^n x_i\cdot 3^{-i} \geqslant h(c),$$
whence
$$f(x) \geqslant \sum_{i=1}^n x_i\cdot 3^{-i} \geqslant h(c).$$
On the other hand, for $x\in W$ we have
$$f(x) \geqslant \sum_{i=1}^{m_2} a_i\cdot 3^{-i} = \sum_{i=1}^{m_2-1} b_i\cdot 3^{-i} + 2\cdot 3^{-m_2} = h(b) + 3^{-m_2} + 3^{-n} > h(b) + 3^{-n},$$
and
$$f(x) \leqslant \sum_{i=1}^{m_0-1} x_i\cdot 3^{-i} + \sum_{i=m_0+1}^\infty 2\cdot 3^{-i} = \sum_{i=1}^{m_0-1} c_i\cdot 3^{-i} + 3^{-m_0} = h(c) - 3^{-m_0} < h(c),$$
and so, since $f$ is bijective, we see that
$$f(W) = C \cap (h(b)+3^{-n}, h(c))$$
is open in $C$. For the excluded cases $a_i = 0$ for all $i$ or $a_i = 2$ for all $i$, the argument uses only one of $b,c$ and leads to $f(W) = C\cap (-\infty,h(c))$ resp. $f(W) = C\cap (h(b)+3^{-n},\infty)$, each of which is open in $C$.
