Let $\{Y_i\}_i$ be the collection of path components of $Y$, so then $X*Y=\bigcup_i X*Y_i$. Let $Z$ be the portion of $X*Y$ associated with $[0,1/2)$, and let $A_i=Z\cup (X*Y_i)$. The intersection of any two of these $A_i$ is $Z$, which is path connected (deformation retracting onto $X$), so if we prove that $X*Y_i$ is simply connected, then $X*Y$ is simply connected by the van Kampen theorem. While it is true that the $A_i$ are not open in $X*Y$, this is fine because the surjectivity part of the theorem only relies on $f^{-1}(A_i)$ being open in $[0,1]$ for any path $f:[0,1]\to X*Y$. We defer showing openness to the appendix.
Because of this, let us assume $Y$ is path connected as well. The space $X*Y$ decomposes into two open subsets, one associated with $[0,2/3)$ and the other associated with $(1/3,1]$. Call these $Z_1$ and $Z_2$. The first deformation retracts onto $X$, the second deformation retracts onto $Y$, and their intersection $Z_1\cap Z_2$ deformation retracts onto $X\times Y$. The inclusion maps $X\times Y\to Z_1$ and $X\times Y\to Z_2$ induce projections on $\pi_1$, so the van Kampen theorem gives $\pi_1(X*Y)$ to be the free product $\pi_1(X)*\pi_1(Y)$ modulo the subgroup generated by all $i_{1*}(x,y)i_{2*}(x,y)^{-1}$, which are all $xy^{-1}$. Since $x,y$ may be arbitrary, the resulting $\pi_1(X*Y)$ is trivial, so $X*Y$ is simply connected.
Appendix. First, $Z$ is open in $X*Y$, so $f^{-1}(Z)$ is open. Take a point $f(s)=(x,y,t)$ in $X*Y_i$ with $t>0$. By continuity of $f$ there is a $\delta$ such that $f((s-\delta,s+\delta))\subset X\times Y\times (0,1]$. Since the image of this interval must lie in a path component, $f((s-\delta,s+\delta))\subset A_i$.
Let $L$ be a half-line. Claim : $U = R^n \backslash L$ and $V = R^n \backslash (-L)$ are simply connected, and $U \cap V = R^n \backslash (L \cup - L)$ is connected, and since $U \cup V = R^n \backslash 0$ we obtain the result.
(In the comments, I took $n=3$ for simplicity but the result holds for any $n \geq 3$.)
Best Answer
The lemma does not apply to your example because $U\cap V=S^1\setminus\{x,y\}$ is not path-connected. Indeed, if $A$ and $B$ denote the two open arcs of the circle from $x$ to $y$, then $A$ and $B$ are both open in $U\cap V$ with $A\cap B=\emptyset$ and $A\cup B=U\cap V$, showing that $U\cap V$ is not connected.