= 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 $B\ne\varnothing$. \b[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 $W$. \b[The union is <path-connected>.] There need not be a single point common to all the subsets.
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