Let $f_i:A_i\rightarrow A_{i+1}$ be a sequence of closed embeddings of topological spaces. Denote $A:=\text{colim}_iA_i$ which is basically a union of all $A_i$ and we say that $U\subset A$ is open (closed) iff $U\cap A_i$ is open (closed) for all $i$. Let $K\subset A$ be a compact subspace. Then there is $i$ such that $K\subset A_i$.
I tried to prove it by contradiction: suppose there is a sequence $x_i\in K$ such that $x_i\notin A_i$. Since $K$ is compact there must be a subsequence $x_{i_n}$ converging to $x\in K\cap A_N$. Funny, but I cannot proceed because open sets seem too large and I can't catch all but finite of $x_i$ in a single $A_i$. I must be missing something simple.
