Conditions for mapping product to be open Let $(E_i,T_i)_{i\in I}$ be a family of topological spaces, and let $(E,T)$ be the product of these spaces. Let $f_i:E\to E_i\ (i\in I)$ be a family of mappings. Suppose that each $f_i$ is $(T,T_i)-$ continuous. 
Denote the product of these mappings by $\displaystyle\prod_{i\in I}f_i,$ or shortly $f.$ 
Denote the image of $f$ by $D,$ and let $T_D$ be the subspace topology on $D$ inherited from $(E,T).$ 
Next we'll introduce a condition for family $ (f_i)_{i\in I}.$

Condition: For any $T-$ closed set $F,$ and for any $x\in E/F,$ there exits some $i\in I,$ such that $f_i(x)\notin\mathrm{Cl}_{T_i}\big(f_i(F)\big).$ (Here $\mathrm{Cl}_{T_i}$ means taking the closure with respect to topology $T_i$.)

Claim: If the family $(f_i)_{i\in I}$ satisfies the condition above, then the map $f$ is  $(T,T_D)-$ open ,i.e. $X$ is a $T-$ open set $\Longrightarrow$ $f(X)$ is a $T_D-$ open set.
Question: Is the claim above true or false? Any help would be appreciated.
Background: I encountered this claim in a topology workbook, A General Topology Workbook by Iain T.Adamson, and I highly suspect its validity. The condition looks quite weird to me. My doubt rose when I found that the proof on that book is wrong. I tried to find one counterexample, but only got lost in details of the construction of each $f_i.$  
Update: I find that in that very book, using the claim above the author proves an astonishing result: every topological space $(E,T)$ is homeomorphic to a subspace of $(\prod_{i\in I}X,\ \prod_{I\in I}T_0),$ here $X=\left\{a,b,c\right\},T_0=\left\{\emptyset,\left\{a\right\},X\right\},\ I=E\cup T.$ 
 A: Let $E_i$ be the set of spaces we take the product of, and $E$ their product (with product topolgoy induced by the projections $\pi_i : E \to E_i$ as usual).
Suppose we have $f_i : X \to E_i$ from some fixed space $X$ (need not be the product $E$ as the OP writes) which obeys the condition that $\{f_i: i \in I\}$ separates points from closed sets, which is the OP's condition:
$$\forall F \subseteq X \text{ closed }: \forall x \in X \setminus F: \exists i\in I: f_i(x) \notin \overline{f_i[F]}$$
where bar denotes closure in $E_i$ (clear from context). The function $f(x)$ from $X$ into $E$ defined by $f(x)_i = f_i(x)$ for all $x \in X, i \in I$ is well-defined and continuous (at least when all $f_i$ are continuous, which is commonly assumed in this setting). 
The condition above is equivalent to the following:

$\mathcal{B}:= \{(f_i)^{-1}(U): i \in I, U \subseteq E_i \text { open }\}$ is a base for the topology of $X$.

which I show here as an answer to a question. 
Now, to my point: knowing that $\mathcal{B}$ is a base, to check $f$ is open from $X$ to $f[X] \subseteq E$ (the set the OP calls $D$) in the subspace topology, we only have to show that $f[O]$ is open in $D$ whenever $O \in \mathcal{B}$ (we only have to check openness on base elements as $f[\cup_i O_i] = \cup_i f[O_i]$ etc.). To see that $f$ sends base elements to open sets: 
let $O = f_i^{-1}[U]$ where $U \subseteq E_i$ is open, so an arbitary element of $\mathcal{B}$. 
Then $$f[O] = D \cap \pi_i^{-1}[U]$$ showing that $f[O]$ is indeed relatively open in $D$. To see the set equality, check two inclusions: if $y \in f[O]$, then certainly $y \in f[X] = D$, but also, $y = f(x)$ with $x \in O= f_i^{-1}[U]$, which implies that $f_i(x) \in U$, but by how $f$ is defined from the $f_i$, this just means $y_i \in U$ or $y \in \pi_i^{-1}[U]$, so indeed $y \in D \cap \pi_i^{-1}[U]$. Reversely, if $y \in D \cap \pi_i^{-1}[U]$ we know that $y_i \in U$ and $y = f(x)$ for some $X \in X$. This $x$ obeys $f_i(x) = f(x)_i = y_i \in U$ so $x \in f_i^{-1}[U] = O$ and so $y \in f[O]$ QED. 
