A subset $U$ of a CW complex is open iff for any closed cell $C$, the set $U \cap C$ is relatively open in $C$. My question is, can one characterize the subspace topology a similar way? That is, (when) is the following true?
For any subspace $X$ of a CW complex, a set $U \subseteq X$ is open in $X$ iff for any closed cell $C$, the set $U \cap C$ is relatively open in $X \cap C$.
This seems a very natural question to me, yet I couldn't find any mention of it. The left-to-right implication is clear from the definition of subpace topology on $X \cap C$.
The reverse implication seems true to me for all metrizable (that is, locally finite) CW complexes, here is a tentative proof: assume by contrapositive that $U \subseteq X$ is not open in $X$, that is there is a sequence $(x_n)_{n \in \mathbb N}$ in $X\setminus U$ with $\lim x_n=x \in U$. Consider the set $S=\{x_n: n \in \mathbb N\} \subseteq X\setminus U$. Since $x \notin S$, $S$ is not a closed set of the complex, so there exists a closed cell $C$ with $S \cap C$ not closed in $C$. Then $S \cap C$ has a limit point in $C\setminus S$, which can only by $x$ as $S$ has no other limit points. But as $S \cap C \subseteq (X\setminus U) \cap C$, the point $x$ is also a limit point of the latter, and $x \in U \cap C$, so $U \cap C$ is not open in $X \cap C$.