Take a point $p$ in $A$ and let $B$ be the set of points of $A$ that can be joined to $p$ by a path within $U_\epsilon(A)$ (not within just $A$!). Then $B$ is easily seen to be open in $A$: if you can join $p$ to $q$, you can join it to all of $U_\epsilon(q) \cap A$. And also closed in $A$: if you can't join $p$ to $q$, it follows you can't join $p$ to any point of $U_\epsilon(q) \cap A$. Since $A$ is connected, $B = A$. But now $U_\epsilon(A)$ is clearly path-connected: any point $x$ of $U_\epsilon(A)$ is within distance $\epsilon$ of some $q \in A$, and we proved you can join $p$ to $q$ (within $U_\epsilon(A)$, and then join $q$ to $x$ by a straight line segment in $U_\epsilon(q)$.
Alternatively: to do it stringing together more standard results, first notice that $U_\epsilon(A)$ is connected since it can be written as a union of connected set with a common intersection, namely, as the union of all $A \cup U_\epsilon(p)$ for $p \in A$. Now, $U_\epsilon(A)$ is connected and locally path connected and therefore connected.