$\pi_1(U\cap V)\cong\mathbb{Z}$ for $U,V$ simply connected Let $X=U\cup V$ where $U,V$ are simply-connected open sets and $U\cap V$ is the disjoint union of two simply connected sets. We also have the condition that any subspace $S$ of $X$ homeomorphic to $[0,1]$ has an open neighborhood that deformation retracts onto $S$.
We can choose points $p$ and $q$, one from each of the two disjoint components of $U\cap V$, that are not connected by a path. Then $U\cup V$ should deformation retract onto the union of two paths (one path $\alpha$ in $U$, another path $\beta$ in $V$) that connects $p$ and $q$, hence the fundamental group must be $\mathbb{Z}$.
But I don't know how to rigorously show this part. We don't know if $U$ deformation retracts onto $\alpha$ and $V$ deformation retracts onto $\beta$.. and even if we show that, I don't know how do we deal with the intersection $U\cap V$.
 A: It seems to me that this  mathoverflow question and answers are relevant here.
The usual assumption is that the answer has to be a group. But there are myriads of connected spaces which arise as open unions of connected spaces  with non connected intersections, in which case it is not clear where to choose a base point. The proposal I made in 1967 was to choose a set say $S$ of base points and prove a van Kampen Theorem for the fundamental groupoid $\pi_1(X,S)$, see a later proof in this paper. Keeping the set $S$ and working with the algebra of groupoids also enables one to keep any information on the symmetry of the situation, as has been strongly argued by Grothendieck in Section 2 of his 1984 "Esquisse d'un Programme". 
May 14: I think people may need more details.
THEOREM:  Let $S$ be a subset of $X= U \cup V$ which meets each path component of $U,V,U \cap V$, where, as in the question,  $U,V$ are open (and nonempty, for convenience).
Then the diagram  of groupoids 
$$\matrix{ \pi_1(U\cap V, S) & \to & \pi_1(V,S) \\
\downarrow && \downarrow\\
\pi_1(U,S)&\to & \pi_1(X,S) }$$
with morphisms induced by inclusions, is a pushout of groupoids. Note that if $Y$ is a subset of $X$, then $\pi_1(Y,S)$ means $\pi_1(Y, S \cap Y)$. 
The proof is by verification of the universal property, so requires no knowledge of how to construct pushouts of groupoids, or even that they exist in general. It is in essence the same as Crowell's 1958 proof for the fundamental group. 
In the case in point we take $S=\{p,q\}$ where $p.q$ are as in the question. Then under the assumptions, the above diagram becomes
$$\matrix{S & \to &  I_S\\
\downarrow & &\downarrow\\
 I_S &\to & \pi_1(X,S)}$$ 
where $I_S$ denotes the groupoid with objects $S$ and only two non identity arrows
$\iota: p \to q$,   $\iota^{-1}: q \to p$. This groupoid seems "trivial", but it plays the same role in the category of groupoids as does the integers $\mathbb Z$  in the category of groups (it is a "generator"in the categorical sense). There is a "universal" morphism   $\nu: I_S \to \mathbb Z$ which identifies the objects $p,q$. This gives a pushout diagram 
$$\matrix{I_S & \to & \{0\}\\
\downarrow && \downarrow\\
\pi_1(X,S) & \to & \mathbb
 Z}$$
which combines with the first pushout to give a pushout which gives the result you want. Actually it gives more as you do not need to assume each component of $U \cap V$ is simply connected. 
Further relevant algebra of groupoids is developed here. 
A: The hypothesis about embedded arcs being neighborhood retracts is unneeded.  Let $W$ and $Z$ be the two path components of $U\cap V$. Build a covering space over $X$ by taking countably many copies $U_i$,$V_i$, $W_i$ and $Z_i$ of $U$, $V$, $W$ and $Z$ respectively each indexed by the integers and gluing each $U_i$ to $V_i$ along $W_i$ and gluing $V_i$ to $U_{i+1}$ along $Z_i$.  The resulting space $\tilde X$ is a simply connected covering space of $X$ and thus is the universal cover of $X$. It is immediate that the deck transformation group of $\tilde X$ over $X$ is given simply by shifting indices and thus is $\mathbb Z$.  This shows that $\pi_1(X)\simeq \mathbb Z$. 
