A metric is continuous I am looking for some feedback on my solution to the following problem. Thank you! :)
Let $\{(X_i, d_i)\mid i = 1,\ldots, n\}$ be a finite collection of metric spaces. We define $d^{*}: \prod^{n}_{i=1} X_i \times \prod^{n}_{i=1} X_i \rightarrow \mathbb{R}$ by the equation $$ d^{*}(\textbf{x},\textbf{y}) = \max \{d_i(x_i,y_i) \mid  i=1, \ldots, n\}$$ for $\textbf{x},\textbf{y} \in \prod^{n}_{i=1} X_i.$
$\textbf{Problem:}$ Let $(X,d)$ be a metric space. Then, $d: X \times X \rightarrow \mathbb{R}$ is continuous. 
$\textbf{Solution:}$ Let $\textbf{x}=(x_0,y_0)$ be an arbitrary point in $X \times X.$ Let $\epsilon > 0$ be given. To show that $d$ is continuous at $\textbf{x}$, we need to find a $\delta > 0$ such that for $\textbf{y}=(x_1, y_1) \in X \times X$ with $d^{*} (\textbf{x}, \textbf{y}) < \delta$, $|d(x_0,y_0) - d(x_1,y_1)| < \epsilon.$ Take $\delta = \frac{\epsilon}{2}.$ Then, if $d^{*}(\textbf{x},\textbf{y})< \delta$, $d(x_0,x_1) < \frac{\epsilon}{2}$ and  $d(y_0,y_1) < \frac{\epsilon}{2}.$ Apply the triangle inequality twice and  obtain the following: \begin{align*}
  d(x_1,y_1) &\leq d(x_1, x_0) + d(x_0, y_1) \\
  &< \frac{\epsilon}{2} + d(x_0, y_1) \\
 &< \frac{\epsilon}{2} + d(x_0, y_0) + d(y_0, y_1) \\
 &< \frac{\epsilon}{2} + d(x_0, y_0) + \frac{\epsilon}{2} \\
 &= d(x_0, y_0) + \epsilon.
\end{align*} Hence, $d(x_1, y_1) < d(x_0, y_0) + \epsilon.$ Similarly, 
\begin{align*}
  d(x_0,y_0) &\leq d(x_0, x_1) + d(x_1, y_0) \\
  &< \frac{\epsilon}{2} + d(x_1, y_0) \\
 &< \frac{\epsilon}{2} + d(x_1, y_1) + d(y_1, y_0) \\
 &< \frac{\epsilon}{2} + d(x_1, y_1) + \frac{\epsilon}{2} \\
 &= d(x_1, y_1) + \epsilon.
\end{align*} This implies $d(x_0, y_0) - \epsilon < d(x_1, y_1).$ Indeed, $|d(x_0,y_0) - d(x_1,y_1)| < \epsilon.$ Because $\textbf{x}=(x_0,x_1)$ was arbitrary, we conclude $d$ is continuous. $\blacksquare$

In the beginning of your proof I think you mean to say take $d(x_{0},x_{1})<\frac{\epsilon}{2}$ and $d(y_{0},y_{1})<\frac{\epsilon}{2}$. Then if $\delta=\frac{\epsilon}{2}$ we have that if $d^{*}(\textbf{x},\textbf{y})<\delta$ then $|d(x_{0},y_{0})-d(x_{1},y_{1})|<\epsilon$. Provided this is what you mean to say I think the rest of the proof is okay.
