Why do we need the subsets to be closed in this theorem? The theorem in question is:
Let $X$ be Hausdorff and $\{F_1, F_2, ..., F_k\}$ a family of closed subsets of $X$. The subspace $F = \bigcup_{i=1}^k F_i$ of $X$ is compact if and only if all subspaces $F_i$ are compact.
I'm having trouble seeing why the $F_i$ need to be closed in this theorem. My rough proof is:
($\rightarrow$) For each $i=1, \ldots, k$ consider an open cover $C_i$ for $F_i$. Taking the unions of these open covers gives us an open cover for $F$. Since $F$ is compact by assumption, we have a finite open subcover for $F$ which yields finite open subcovers for each $F_i$ by using only the subcover sets that intersect each $F_i$.
I won't go into the other direction, as I have a feeling only this direction required the closed hypothsesis. Perhaps I'm wrong?
 A: If $A$ and $B$ are two compact subsets of a space $X$, then their union $A\cup B$ is also compact, without requiring $X$ to be Hausdorff or $A$ and $B$ closed. You need being Hausdorff of $X$ and closedness of the sets $F_i$ to prove the other direction.
A: Your proof is incorrect.  If you want to prove $F_1$ is compact, you need to start with an arbitrary open cover $C_1$ of $F_1$ and find a finite subcover of it.  Your argument then chooses open covers $C_2,\dots,C_n$ of $F_2,\dots,F_n$ and finds a finite subcover $D\subseteq\bigcup C_i$.  However, $D$ is probably not a subset of $C_1$, so it does not give a finite subcover of $C_1$!  Even if you restrict to those elements of $D$ which intersect $F_1$, some of them may come from $C_i$ for $i>1$ rather than from $C_1$.
For an explicit counterexample, consider that $[0,1]$ is compact and is the union of any subset $C_1\subseteq [0,1]$ and $C_2=[0,1]$ no matter what $C_1$ is.  So your argument would prove that every subset of $[0,1]$ is compact, which is certainly false.
(On the other hand, the assumption that the $F_i$ are closed is not needed for the reverse direction.  Indeed, in any space, any finite union of compact subsets is compact.  Incidentally, the assumption that $X$ is Hausdorff is also unnecessary for both directions.)
