Show $ \{ (\xi,\eta,\zeta) \in \mathbb{R^3} : \xi = \eta = \zeta \}$ is closed Show $ F =\{ (x_1,x_2,x_3) \in \mathbb{R^3} : x_1 = x_2 = x_3 \}$ is closed.
I'd like help finishing off my solution below. Other answers are appreciated as well.

It suffices to show that the complement  $F^C$ is open.
Now the complement can be written as follows
$$
\begin{align}
F^C 
&= \{ (x_1,x_2,x_3) \in \mathbb{R^3} : \neg( x_1 = x_2 = x_3)\}\\
&= \bigcup_{i< j} \{  (x_1,x_2,x_3) \in \mathbb{R^3} : x_i \not= x_j \}\\
&= \bigcup_{i< j}\left(  \{  (x_1,x_2,x_3) \in \mathbb{R^3} : x_i > x_j \} \cup \{  (x_1,x_2,x_3) \in \mathbb{R^3} : x_i < x_j \} \right)
\end{align}
$$ 
Now the union of a collection of open sets is open.
Assume without loss of generality $i=1, \ \ j=2$, it then suffices to show that  $ A \ \ = \{  (x_1,x_2,x_3) \in \mathbb{R^3} : x_1 < x_2 \}$ is an open set in $\mathbb{R^3}$. (The proof that $ A^* = \{  (x_1,x_2,x_3) \in \mathbb{R^3} : x_1 > x_2 \}$ is an open set hopefully being very similar.)
To this end:
Fix $a = (a_1,a_2,a_3) \in A$
Choose $r = |a_1 - a_2|$
Fix $b = (b_1,b_2,b_3) \in \mathbb{R^3}$
Is it true that $\| a - b \| < r \Rightarrow b \in A \ \ ?$ 
If so, how and why? If not what is a suitable choice for r?

Reference
This was Example 8.5.(e) taken from:
Bartle, Robert G., The Elements of Real Analaysis, John Wiley and Sons, 1964
 A: Define $f(x) = (x_1-x_2)^2+(x_2-x_3)^2$. Then $F=f^{-1} (\{0\})$ is closed because $f$ is continuous and $\{0\}$ is closed.
Similar way: Let $A=\begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1\end{bmatrix}$. Then
$F = \ker A$ which is closed since $x \mapsto Ax$ is continuous.
Another, slightly different way. Suppose $x_n \to x$ and $x_n \in F$ for all $n$. Then each component converges to the same value (since $[x_n]_k = [x_n]_1$ for all appropriate $n,k$). Hence $x=([x]_1,[x]_1,[x]_1)$ and so $x \in F$.
A: I believe there are quicker, more abstract approaches (I think you should be able to exploit the fact that projections are open maps to reduce quickly to $\mathbb R$).
But, to conclude your proof, you need the following: given $a \in A$, there exists an $\epsilon>0$ such that $\Vert b-a \Vert < \epsilon$ implies $b \in A$, i.e. that $b_1 < b_2$.
Notice that $\Vert b -a \Vert^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2 + (a_3 - b_3)^2 \ge \max_i \vert a_i - b_i \vert ^2$.
Let $r = a_2 - a_1 > 0 $. Then
$$ b_1 - b_2 
\le \vert b_1 - a_1 \vert + \vert b_2 - a_2 \vert + a_1 - a_2  \le 2 \Vert b - a\Vert - r {}{}{}{} < 2 \epsilon - r $$
So that taking $\epsilon = r/2$ does the job.
A: Define $$f:\mathbb{R}^3\to\mathbb{R}\quad f(x_1,x_2,x_3)=x_2-x_1$$. Observe that $$A=\{(x_1,x_2,x_3):\mathbb R^3:f(x_1,x_2,x_3)\in(-\infty,0)\}$$ $(-\infty,0)$ is an open set in $\mathbb{R}$ and since this function is continuous it finishes the proof. Do the same for the other cases. About your claim: maybe it's correct but usually we tend to prefer choosing a contiguous function where the source of every open set is open (and same for close sets).
