I just need help understanding this problem:

Suppose that $\displaystyle \{X_i,\Omega_i) : i\in I\}$ is a family of topological spaces and suppose that $\displaystyle \bold{x}$ and $\displaystyle \bold{y}$ are members of $\displaystyle \prod \{X_i : i\in I\}$. If $\displaystyle \bold{x}$ and $\displaystyle \bold{y}$ differ in only one coordinate, prove that they must lie in the same connected component.

What we have to work with:

Let $\displaystyle (X,\Omega)$ and $\displaystyle (Y,\Theta)$ be topological spaces. If $\displaystyle (X,\Omega)$ is connected and the function $\displaystyle f: X \to Y$ is continuous relative to $\displaystyle \Omega$ and $\displaystyle \Theta$, then $\displaystyle f(X)$ is connected as a subspace of $\displaystyle (Y,\Theta)$.

Suppose that $\displaystyle \{X_i,\Omega_i) : i\in I\}$ is a family of topological spaces. Fix $\displaystyle i\in I$; and for all $\displaystyle j\in I - \{i\}$, fix $\displaystyle a_j \in X_j$. Define the function $\displaystyle f_i: X_i \to \prod\{X_i : i\in I\}$ by $\displaystyle f_i(x) = \bold{x}$, where $\displaystyle \pi_i(\bold{x}) = x$ and $\displaystyle \pi_j(\bold{x}) = a_j$ when $\displaystyle j\neq i$. Using this map, it can be shown that $\displaystyle (X_i,\Omega_i)$ is homeomorphic to the subspace $\displaystyle f_i(X_i)$.

Both of these have been proven in class. I'm just failing to see how to use these when one coordinate differs. Any help is greatly appreciated!