Prove that a closed rectangle in $\mathbb{R^n}$ is a closed set. Prove that a closed rectangle in $\mathbb{R^n}$ is a closed set.
My trial:
Define the closed rectangle  by: $A=\{a_{1} \leq x_{1} \leq b_{1},a_{2} \leq x_{2} \leq b_{2}, ...,a_{n} \leq x_{n} \leq b_{n} \}$
A closed set is defined as a set that contains all its cluster points.
Distinguish between 2 cases:
Case1: $x \in A$.
Then there exist an interval $[a,b]$ such that $a \leq x \leq b$ by the definition of a closed rectangle and then $A \cap [a,b]$\ $\{x\} \neq \phi$.
Is this justification correct?
Case2:$x \notin A.$
Then distinguish between 2 cases:
Case a: 
$\inf|| x - A|| = 0$ 
Then in this case $x$ is a cluster point of A (either A is closed or open it does not matter) then there exist a sequence $c_{k} \in A$ such that $c_{k} \rightarrow x$ as $k \rightarrow \infty.$ But since $c_{k} \in A$ then $a_{i} \leq c_{k} \leq b_{i}.$ Then by squeeze theorem and since  $a_{i}$ & $b_{i}$ are constants we have $a_{i} \leq x  \leq b_{i}.$ which means that $x \in A$ which is a contradiction to our assumption..... but I do not know what should I conclude then? ...... may be this case can not exist i.e. the infimum can not be equal zero?
Case b:
$\inf||x - A|| \neq 0 $ 
Then put $ a = \inf||x - A|| ,$ then the interval $[x - a, x + a]$ does not contain other points of $A$ and hence $x$ is not a cluster point of $A$ in this case..... is my argument correct in this case?
Could anyone help me filling the gaps in my proof please?
EDIT 1:
Case1: $x \in A$.
Then there exist $1 \leq i \leq n$ such that $a_{i} \leq x_{i} \leq b_{i}$ by the definition of a closed rectangle and then $A \cap [a_{i},b_{i}]$\ $\{x_{i}\} \neq \phi  \forall  i$.
And hence $x_{i}$ is a cluster point of $A$ $\forall i$ and $x \in A$ ...... is my new edit correct?
EDIT 2:
Case b:
The following case should be deleted because it is not a cluster point.
$\inf||x - A|| \neq 0 $ 
Then put $ a = \inf||x - A|| ,$ then the interval $[x - a, x + a]$ does not contain other points of $A$ and hence $x$ is not a cluster point of $A$ in this case..... is my argument correct in this case?
 A: I cannot comment yet so I must post in answer. Sorry about that. 
Well any $[a_i,b_i]$ is closed in $\mathbb{R}$. Now the Cartesian product $A \times B$ of two closed set $A$ (closed in $X$) and B (closed in $Y$) is closed in $X \times Y$ (see Is product of two closed sets closed?). So then your rectangle is closed in $\mathbb{R}^n$ as the Cartesian product of $n$ closed sets in $\mathbb{R}$.
A: I'll try though I prefer Plussoyeur's answer. Consider $x\in\mathbb{R}^n$ so that for every neighborhood $U\in \mathcal{U}(x)$, ( the system of neighborhoods) You have
$$U\cap A\neq\emptyset.$$
(So $x$ is an  clusterpoint of $A$.) Call $$P_i:\mathbb{R}^n\rightarrow\mathbb{R},\begin{pmatrix}x_1\\ \cdot\\\cdot\\x_i\\\cdot\\x_n\end{pmatrix}\mapsto x_i$$
the projection on the $i$'th component and for each $U\in\mathcal{U}(x)$ define $U_i=P_i(U)$ for $i=1,...,n$. Now by the product-topology the projections are open and thus $U_i$ is a neighborhood of $x_i$ that satisfies
$$U_i\cap[a_i,b_i]\neq\emptyset$$
for $i=1,...,n$. Thus $x_i\in[a_i,b_i]$ for $i=1,...,n$ and so $x\in A$.
