The compactness argument is quite valid, $(0,1)^2$ is not closed so not compact and the continuous image of a compact space is compact.
The completeness argument is not correct. The continuous image of a complete metric space (even within the space) need not be complete, e.g. $\arctan(x): \Bbb R \to \Bbb R$ has image $(-\frac{\pi}{2}, \frac{\pi}{2})$. Completeness is a property of the metric, not of the topology and continuous functions need not preserve metric values (isometries do).
There are also more advanced topological properties that the closed square has and the open one does not, but I think compactness is the simplest solution to this problem.
Suppose we limit our language drastically by only allowing ourselves to refer to sets and subsethood-comparability; so we can't say "$X$ has a single element" or "$X\subseteq Y$," but we can say "Either $X\subseteq Y$ or $Y\subseteq X$" (denote this latter relationship by $\gtreqless$ for simplicity). Then for any topological space $\mathscr{A}=(A;\tau)$, the resulting "set comparability structures" $$Open_\mathscr{A}=(\tau; \gtreqless)\quad\mbox{and}\quad Closed_\mathscr{A}=(\check{\tau}; \gtreqless)$$ (where $\check{\tau}=\{U\subseteq A: A\setminus U\in \tau\}$ is the set of $\tau$-closed sets) are isomorphic via the complement map $U\mapsto A\setminus U$. So any way of distinguishing them must involve either referring to points (= elements of $A$) directly or making a distinction between $\subseteq$ and $\supseteq$. This seems to match the notion of duality you have in mind, and so we get a positive answer to at least one version of your question.
Once we can count points or refer to $\subseteq$ as such, however, open and closed are quite easily distinguishable. For example, the statement "If every singleton is closed, then every set is closed" is not true in all metric spaces, but the statement "If every singleton is open, then every set is open" is true in all metric spaces. Relatedly, consider "There exist minimal nonempty closed sets" (true in all metric, or even $T_1$, spaces) versus "There exist minimal nonempty open sets" (false even in the usual topology on the real line).
Narratively, I'd say that the real question is whether we think in terms of sets or in terms of ways of cutting the space. The former lets us distinguish closed-ness and open-ness in most settings, but the latter is much coarser; in particular, an open set $U$ corresponds to literally the same "cut" - namely, $A=U\sqcup (A\setminus U)$ - as does its closed complement $A\setminus U$. (I've talked about a similar focus-shift at the end of this old answer, in the context of symmetry in logic with respect to negation; look for the phrase "assertion logic," and don't be too put off by the technical character of the rest of the answer.)
Best Answer
For 1) take a constant function and let it be that singletons in the codomain are not open.
For 2) take the identity $\mathbb R\to\mathbb R$ where codomain is discrete (so $d(x,y)=0$ if $x=y$ and $d(x,y)=1$ otherwise) and domain is not discrete.