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Given that $(X_1, d_1), (X_2, d_2), \ldots, (X_n, d_n)$ are metric spaces and $X = X_1 \times X_2 \times \cdots \times X_n$ is converted into a metric space $(X, d)$ in the where $d((x_1,x_2,...,x_n),(y_1,y_2,...,y_n))=\underset{i\in\lbrace 1,...,n\rbrace}{\max}d_i(x_i,y_i)$, the task is to prove that an open ball in $(X,d)$ is the product of open balls from $X_1, X_2, \ldots, X_n$, respectively.

I am a visual person, but I am having trouble visualizing what this is really saying. Frankly, I'm not even sure what this is asking me to prove beyond notation. All I can muster is that we're trying to show that $$B(x;\varepsilon) = B(x_1;\varepsilon_1) \times B(x_2;\varepsilon_2) \times \cdots \times B(x_2;\varepsilon_n)$$where $x\in X$ and $x_1 \in X_1, x_2\in X_2, \ldots, x_n \in X_n$ and $\varepsilon, \varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n > 0$.

Since $x \in X$, we have $x = (x_1, x_2, \ldots, x_n)$... but that's as far as I can go. Any hint as to how to proceed or (honestly) what it is we're really trying to show would be greatly appreciated.

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    My hunch would be to deal with what seems to be an issue, with the fact that the different metric spaces have different metrics? The way I visualize this is to think about 2D, distance being measured in cm along one dimension and in mm along another one. In terms of notation, you need to argue that for any $\epsilon$ in $(X,d)$, you can find 'corresponding' $\epsilon_{i}$ in $(X_{i},d_{i})$? – 2017-02-16
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    Thank you for the visualization technique! But... I'm still trying to understand what this means, so I've decided to try showing set equality by showing that each is a subset of the other. However, I've run into an issue when I let $a \in B(x,\varepsilon)$. From that assumption, we see that $d((a_1,a_2,\ldots,a_n),(x_1,x_2,\ldots,x_n)) < \epsilon$. How do we go from there to $d(a_1,x_1)<\varepsilon_1, \ldots, d(a_n,x_n)<\varepsilon_n$? – 2017-02-16
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    Having thought about the problem, I am skeptical about the result being correct (or maybe there is something I do not understand). Here is an example: $(X_{1},d_{1})$ and $(X_{2},d_{2})$ are both real lines with the standard Euclidean metric. Then open ball in each is an interval. Product of two intervals is a square. But your $X$ is $\mathbb{R}^{2}$ with open ball being a circle. What am I missing? – 2017-02-16
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    The metric defined on $X$ is $\max_{1\leq i \leq n}\{d_i(x_i,y_i)\}$. Does that change anything? I'm not seeing why the open ball on $X$ is a circle. – 2017-02-16
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    Indeed, this is crucial (and was omitted from your original question). With this, I believe you are done if you choose $\epsilon_{i}=\epsilon$ in your original formulation? – 2017-02-16
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    Apologies--my text used the phrase "in the standard manner," so I had assumed that it was standard outside my text (hence me leaving it out). – 2017-02-16

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Claim: Given $x=(x_1,x_2,...,x_n)$ and $\epsilon>0$, then $$B_X(x,\epsilon)=\underset{i\in\lbrace 1,...,n\rbrace}{\Pi}B_{X_i}(x_i,\epsilon)$$

Proof: $y=(y_1,...y_n)\in B_X(x,\epsilon)\iff d(x,y)=\underset{i\in\lbrace 1,...,n\rbrace}{max}d_i(x_i,y_i)<\epsilon\iff\forall i\in\lbrace 1,...,n\rbrace,d_i(x_i,y_i)<\epsilon\iff\forall i\in\lbrace 1,...,n\rbrace,y_i\in B_{X_i}(x_i,\epsilon)\iff y\in\underset{i\in\lbrace 1,...,n\rbrace}{\Pi}B_{X_i}(x_i,\epsilon)$