Prove that there exists $\delta > 0$ such that $F(x,y)= \|f(x)-f(y)\| < \epsilon$ I am working on some exam practice questions, and this one seems quite hard. I would greatly appreciate some help with it.
Here is the complete question.

Let $ B \subseteq \mathbb{R^d} $ be a bounded and closed set, and let
  $f:B \rightarrow \mathbb{R^N}$ be continous. 
Set $B \times B := \{(x,y) : x,y \in B \} \subseteq \mathbb{R^d} \times  \mathbb{R^d} = \mathbb{R^{2d}} $
   and define $F,h: B \times B \rightarrow \mathbb{R} $
   by $$F(x,y):= ||f(x)-f(y)|| $$ and 
  $$h(x,y):= ||x-y||$$ for all $(x,y) \in B \times B$. The functions $F$
  and $h$ are continuous functions (do not require proof).
Fix $\epsilon > 0$. By considering the continuous function $h$ on the
  set $F^{-1}[\space[\epsilon,\infty)\space]$,
Prove that there exists $\delta > 0$ such that $$F(x,y)= ||f(x)-f(y)||<\epsilon $$
   for all $(x,y) \in B \times B$ such that $ h(x,y)=||x-y|| < \delta$.
   (By the definition of continuity this means that
  $\delta$ can be chosen independently of $x \in B$).

Here is my very humble attempt to start this question:
Let $\epsilon > 0$ be fixed, such that $$F(x,y)= ||f(x)-f(y)||<\epsilon $$ with $x,y \in [\epsilon, \infty) \in B \times B$
Then by the definition of continuity, there exists $\delta > 0 $ such that $ ||x-y|| < \delta$ 
Now consider $h$ on the set $F^{-1}[\space[\epsilon,\infty)\space]$. Then $ h(x,y)=||x-y|| < \delta$ for the same $x,y$ as above...
Now I'm not sure where to go with it. I know it's very far from correct, so would greatly appreciate some help!
Many thanks in advance!
 A: Assume by contradiction this is not true. 
Then for each $n$ you can find some $x_n,y_n$ such that 
$$F(x_n,y_n) > \epsilon \,,$$
$$h(x_n,y_n) <\frac{1}{n} \,.$$
Now, since $B$ is bounded an closed, $x_n$ must have a convergent subsequence 
$$x_{k_n} \to x \,.$$
Same way $y_{k_n}$ has a convergent subsequence $y_{l_n} \to y$. Now 
$$(x_{l_n},y_{l_n}) \to (x,y)$$
and $F,h$ are continuous. Then you get a contradiction by applying the limits to the above inequalities.
Second proof
As $F$ is continuous, $F^{-1}[\space[\epsilon,\infty)\space]$ is closed in $B$ hence $F^{-1}[\space[\epsilon,\infty)\space]$ is closed and bounded in $R^n$.
Therefore, as $h$ is continuous, the statement below you covered is true:


*

*$h(F^{-1}[\space[\epsilon,\infty)\space])$ is compact in $R$.

*$h(F^{-1}[\space[\epsilon,\infty)\space])$ is closed and bounded  in $R$.

*$h: F^{-1}[\space[\epsilon,\infty)\space] \to R$ attains its absolute minimum.


Now, it is easy to prove that $0 \notin h: F^{-1}[\space[\epsilon,\infty)\space]$. Therefore, any of the three statements above imply that there exists some $\delta$ such that $(0, \delta) \cap h(F^{-1}[\space[\epsilon,\infty)\space]) =\emptyset$, which is exactly what you need to prove.
