Showing that the union of three pairwise intersecting sets is not contractible Let $X$ be a "nice" Hausdorff space (probably a compact complete metric space).
Suppose that $A_1,A_2$ and $A_3$ are compact, path-connected subsets of $X$ such that:


*

*$A_i \cap A_j \ne \varnothing$ is path-connected for all $i,j$; and

*$A_1 \cap A_2 \cap A_3 = \varnothing$.


Is the space $A_1 \cup A_2 \cup A_3$ not contractible, and if so how would you go about showing it?
Also, what are the minimum conditions that can be put on "nice"?
 A: It suffices to assume that $X$ is normal (which is stronger than Hausdorff but weaker than metrizable). In particular, any two disjoint closed subsets can be separated by a continuous function (Urysohn's Lemma). Furthermore, you do not need the subsets $A_i$ to be compact, only closed; also, you do not need $A_i\cap A_j$ to be path-connected, only nonempty. I will write a proof as a sequence of lemmas, let me know if you have hard time proving any of these.
Assumption: $X$ is normal, $A_1, A_2, A_3$ are closed, nonempty, path-connected, such that $A_i\cap A_j\ne \emptyset$ for all $1\le i, j\le 3$ and
$$\
A_1\cap A_2 \cap A_3=\emptyset. 
$$
Lemma 1. Given three closed subsets $A_1, A_2, A_3$, as above, 
$$
\exists~ a_i\in A_i - (A_j \cup A_k)
$$ 
for any triple $\{i, j, k\}=\{1, 2, 3\}$.
Lemma 2. There exist open neighborhoods $U_1, U_2, U_3$ of the subsets $A_1, A_2, A_3$ in $X$ such that
$$
U_1\cap U_2 \cap U_3=\emptyset. 
$$
Lemma 3. (Urysohn) There exist non-negative continuous functions $\chi_i$, on $X$ such that $\chi_i=1$ on $A_i$ and $\chi_i=0$ on $X- U_i$, $i=1, 2, 3$. 
Given these functions, I will define a partition of unity on 
$$
A= A_1\cup A_2 \cup A_3, 
$$
$$
\eta_i= \frac{\chi_i}{\chi_1 + \chi_2 + \chi_3}, 
$$
subordinate to the covering of $A$ by the subsets $U_i\cap A_i$, $i=1, 2, 3$. 
The key fact is that the nerve of the above covering is isomorphic to the boundary (topological circle) $S$ of the standard 2-dimensional simplex
$$
\Delta= \{(t_1, t_2, t_3)\in {\mathbb R}_{\ge 0}^3: t_1+t_2+t_3=1\}.
$$ 
(If you do not know what a "nerve" is, just think of this sentence as defining $S$ and $\Delta$.) 
Define the "map-to-nerve" map
$$
f: A\to \Delta, \quad f(a)= (\eta_1, \eta_2, \eta_3). 
$$
The image of this map is contained in  $S$ and
$$
f(a_1)= b_1=(1, 0, 0), \ f(a_2)=b_2=(0,1,0),\ f(a_3)=b_3=(0, 0, 1). 
$$
In addition to the map $f$ we define a map $g: S\to A$ which sends each $b_i$ to $a_i$ and whose restriction to each line segment $[b_i, b_{i+1}]$ ($i$ is taken modulo $3$) is a path in  $A_i\cup A_{i+1}$ connecting $a_i$ to $a_{i+1}$. Here I am using path-connectivity of $A_i\cup A_{i+1}$. 
Lemma 4. The composition $h:= f\circ g: S\to S$ is homotopic to the identity map. 
(Hint: Use the fact that the restriction of $h$ to each segment $[b_{i}, b_{i+1}]$ fixes the boundary of this segment and its image is disjoint from $b_{i-1}$.) 
We are now ready to prove: 
Theorem. $\pi_1(A)\ne \{1\}$, in particular, $A$ is not contractible. 
Proof. I claim that the map $g: S\to A$ is not null-homotopic. Suppose to the contrary that this map extends to a continuous map $G: \Delta\to A$. The composition $f\circ G: \Delta\to S$ is homotopic to the identity when restricted to $S$. But this would imply that $\pi_1(S)=1$, which is a contradiction. qed  
In fact, you do not need path-connectivity of $A_i$'s, only connectivity. But a proof would require Chech cohomology which you probably are not familiar with. 
Edit. Here is a sketch of an argument which assumes mere connectivity of the sets $A_i$, instead of path-connectivity. I will show that 
$$
\check{H}_1(A)\ne 0.
$$ 
One defines Chech homology groups of a topological space $T$ as the inverse limit
$$
\lim_{\leftarrow} H_1(C_\gamma)
$$
where $C_\gamma$'s are nerves of open covers of  $T$ where the order is given by the refinement relation of the open covers. The simplicial complexes $C_\gamma(A_i)$ associated with the sets $A_i$ are not just connected but path-connected. Therefore, instead of constructing maps of intervals to $A_i\cup A_j$  as in the above proof, one constructs maps to the nerves of the open covers. As above, one verifies that all the maps
$$
S\to C_\gamma(A)
$$
are not null-homologous. However, more care (in choosing the maps $S\to C_\gamma(A)$) is needed to ensure that these maps define an element of the inverse limit
$$
\lim_{\leftarrow} H_1(C_\gamma(A)).
$$
In the end, you prove that $\check{H}_1(A)\ne 0$. Since the Chech homology is homotopy-invariant, it follows that $A$ is non-contractible. 
One can also argue using the Chech fundamental group or the Chech cohomology, whatever is feel more comfortable with. 
