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Let $X$ be a CW complex with cell-partition $mathcalC subset mathcalP(X)$, and $mathcalD subset mathcalC$. Is $A = bigcup overlineD(X) : D in mathcalD$ closed in $X$, where $overlineD(X)$ is the closure of $D$ in $X$?

I think I can prove the result when $X$ is first-countable as follows. Suppose $A$ is not closed in $X$. Let $x in overlineA(X) setminus A$. Since $X$ is first-countable, there exists a decreasing neighborhood basis $(U_n in mathcalT_X(x) : n in mathbbN)$ at $x$, where $mathcalT_X(x)$ is the set of neighborhoods of $x$. Let $D_n in mathcalD$ be such that $D_n cap U_n neq emptyset$ and $D_n cap bigcup overlineD_m(X) : m < n = emptyset$. Let $x_n in D_n$. Then $x_n to x$. Since $X$ is Hausdorff, $x_n = overlineD_n(X) cap x_n$ is closed in $overlineD_n(X)$. Since $mathcalT_X$ is coherent with $mathcalC$, $B = x_n : n in mathbbN$ is closed in $X$, which contradicts $x in overlineB(X) setminus B$.

Can this result be proved without assuming first-countability?

EDIT: Seems like the proof above is incomplete. It should also show that $overlineC(X) cap B$ is closed in $X$ for all $C in mathcalC setminus mathcalD$. Following Eric's suggestion below, since $X$ is closure-finite, $overlineC(X)$ intersects only finitely many cells in $mathcalD$, which means $overlineC(X) cap B$ is finite and so closed in $overlineC(X)$. After this, coherence can be applied, which completes the proof.

POST-ACCEPT EDIT: This result is also true for normal CW complexes, which is where the closure of each cell is a sub-complex. This is because sub-complexes are closed under arbitrary unions and intersections. The counter-example given in the accepted answer is neither locally finite or normal, as required. This and other results can be found from the book "The Topology of CW Complexes".

This is not true in general. For instance, let $Asubset S^1$ be any non-closed subset. For each $ain A$, attach a $2$-cell to $S^1$ via the constant map $S^1to S^1$ with value $a$ and let $X$ be the resulting CW complex. Then the union of the closures of all the $2$-cells in $X$ is not closed, since its intersection with $S^1$ is $A$.

(Incidentally, your proof in the first-countable case is not quite correct, since you have not actually shown that $B$ has closed intersection with each closed cell; you only checked this for the cells $D_n$. To prove it for an arbitrary closed cell, you need to use the fact that a closed cell can only intersect finitely many of the $D_n$.)