Is a topological space $X$ the colimit of an open cover $\cup U_i$ in this way? Let $X$ be a topological space space and $X=\cup_{i\in I} U_i$ a covering of $X$ by open subsets $U_i\subseteq X$.

Is it true that
  $$
\operatorname{colim}\left(\coprod_{(i,j)\in I\times I} U_i\cap U_j \rightrightarrows\coprod_{i\in I} U_i\right) \cong X
$$
  where the two maps are the inclusions of either $U_i\cap U_j$ into $U_i$ or $U_i\cap U_j$ into $U_j$? If this is not true in general, is it at least true for $X$ a CW-complex and $\cup_{i=0}^n U_i$ a finite good cover?

 A: The nice concept to use here is that of coequaliser. Thus the diagram 
$$\coprod_{(i,j)\in I\times I} U_i\cap U_j \rightrightarrows^a_b\coprod_{i\in I} U_i \to^c  X$$ 
is a coequaliser  in the sense that a map $f: X \to Y$ is completely determined by maps $f_i: U_i \to Y, i \in I$ such that $f_ia=f_ib$ for all $i$. The condition of an open cover is sufficient for this. The nice thing is that  you get an analogous coequaliser on fundamental groupoids: 
$$\coprod_{(i,j)\in I\times I}\pi_1( U_i\cap U_j)  \rightrightarrows^a_b\coprod_{i\in I} \pi_1(U_i) \to^c  \pi_1(X).$$ 
A: This is true in complete generality by just checking the universal property of the colimit (which is in your case also called a coequalizer, as already mentioned by Ronnie Brown):
A cocone $f$ over the diagram of your question gives a family $\{f_i\colon U_i\rightarrow Y\}_{i\in I}$ of continuous maps, such that that the restrictions $f_i|_{U_i\cap U_j}=f_j|_{U_i\cap U_j}$ coincide. This means we get a unique well-defined map of sets $X\rightarrow Y$ in a way that everything commutes. This map is also continuous by the pasting lemma, which proves the universal property of the colimit.
