 Solution

Suppose the union $W$ has a separation $W=A\cup B$ into disjoint, nonempty, relatively open subsets. Each connected $U_j$ must lie entirely in one of them: otherwise its intersections with $A$ and $B$ would separate $U_j$ in its subspace topology. In particular $U_1$ lies on one side, say $A$. Every other $U_j$ intersects $U_1$, so it cannot lie in $B$ and must also lie in $A$. Then $W\subseteq A$, contradicting $B\ne \varnothing$. The union is connected. This is the connected union with a connected hub argument.

For path-connectedness, take any $x\in U_i$ and $y\in U_j$. Choose $a\in U_i\cap U_1$ and $b\in U_j\cap U_1$. There is a path from $x$ to $a$ inside $U_i$, a path from $a$ to $b$ inside $U_1$, and a path from $b$ to $y$ inside $U_j$. Concatenate the three paths on successive thirds of $[0,1]$. They agree at the joining points, so the resulting map is continuous and remains in $W$. The union is path-connected. There need not be a single point common to all the subsets.