Prob. 2(b), Sec. 25, in Munkres' TOPOLOGY, 2nd ed: The iff-condition for two points to be in the same component of $\mathbb{R}^\omega$
What you did seems great.
Some comments:
- A hint was provided in the exercise, which is that it is sufficient to consider the case where one of the sequence is always vanishing. Indeed it simplifies the notations. You can prove that considering this case is enough by considering the translation $\tau_a$ such that $\tau_a(\mathbf 0)= \mathbf a$. $\tau_a$ is an homeomorphism of $\mathbb R^\omega$, and therefore transform components into components.
- Regarding your question to avoid using path connectedness. You can notice that the sup norm is well defined on $U$. Also, the open ball $B$ centered on $\mathbf a$ with radius $2\Vert \mathbf a -\mathbf b\Vert_\infty$ is open, connected as all open balls and included in $U$. Therefore $U$ is connected. This is a general result that a union of connected sets that all contain a point $x$ is connected.