Proof:

Let us choose a base point $x_0\in U\cap V$ and $\gamma$ be an arbitrary loop in $X$ based at $x_0$. The open cover $\{\gamma^{-1}(U), \gamma^{-1}(V)\}$ of $[0, 1]$ has a Lebesgue number $\epsilon$. Choose a partition

$$\{t_0 = 0 < t_1 < t_2 < \dots < t_n = 1\}.$$

such that the length of each sub-interval is less than $\epsilon$. Then $\gamma$ maps each $[t_j, t_{j+1}]$ into $U$ or $V$. If $\gamma$ maps two adjacent intervals into $U$ or into $V$ then drop the abutting point of the two intervals thereby coarsening the partition. Thus may arrange it such that for each $j = 1, 2,\dots, n-1$, the point $\gamma(t_j)$ lies in $U\cap V$. We now choose a path $\sigma_j$ joining $x_0$ and $\gamma(t_j)$ such that the image of $\sigma_j$ lies entirely in $U\cap V$. This is possible since $U\cap V$ is path connected and $x_0\in U\cap V$. Also let $\gamma_j$ denote the restriction of $\gamma$ to the sub-interval $[t_{j-1}, t_j]$ ( $j = 1, 2, \dots, n$). We may reparametrize $\gamma_j$ (retaining the name) so that its domain is $[0, 1]$. Now

$$\gamma \sim \gamma_1*\sigma_1^{-1}*\sigma_1*\gamma_2*\sigma_2^{-1}*\sigma_2*\gamma_3*\dots*\sigma_{n-1}^{-1}*\sigma_{n-1}*\gamma_n$$

Now each of the loops $\gamma_1*\sigma_1^{-1}$, $\sigma_1*\gamma_2*\sigma_2^{-1}$,$\dots$, $\sigma_{n-1}*\gamma_n$ based at $x_0$ lies in one of the simply connected open sets $U$ or $V$ and so each of them is homotopic to the constant loop via a homotopy $F_j$. These homotopies $F_j$ may be juxtaposed to provide a homotopy between

$$\gamma_1*\sigma_1^{-1}*\sigma_1*\gamma_2*\sigma_2^{-1}*\sigma_2*\gamma_3*\dots*\sigma_{n-1}^{-1}*\sigma_{n-1}*\gamma_n$$

and the constant loop. The proof is complete.