Equivalence statement for a metric space; Munkres chapter 35 question 3 In Munkres's Topology is the following question:
Let $X$ be a metric space, the following are equivalent:


*

*$X$ is bounded under every metric that gives the topology of $X$.

*Every continuous function $\phi: X \to \mathbb{R}$ is bounded.

*$X$ is limit point compact.
I need help proving 2. using 1.
The clue he gives is that if $\phi: X \to \mathbb{R}$ is continuous then $x \mapsto (x,\phi(x))$ is an imbedding.
But the image $\phi(X)$ may not necessarily bounded, can you give me a hint for this?
For 2 implies 3:
Let $A \subset X$ be infinite without a limit point.
$\forall a \in A$ $\exists a \in U$ open in $X$ s.t $U \cap A = \{a\}$. Hence $A$ has the discrete topology as a subspace. Then any surjection $\phi$ from $A$ onto $\mathbb{N}$ is continuous. Define $j \circ \phi: A \to \mathbb{R}$ with $j$ the inclusion map, it is continuous and hence bounded by 2 in  contradiction to it being onto $\mathbb{N}$.
For 3 implies 1:
In a metric space $X$ being limit point compact is saying $X$ is compact, and so bounded.
 A: That 3. implies 1. can be more direct: Suppose that $X$ is limit point compact but unbounded in a compatible metric $(X,d)$. 
Then there is for any $x_0 \in X$ a sequence $(x_n)_n \subseteq X$ such that $d(x_0,x_n) > n$ for all $n$, as $X \nsubseteq B(x_0, n)$.
This set $\{x_n: n \in \mathbb{N} \}$ has a limit point $p$, which implies (in a metric space) that any $B(p, r)$, $r>0$ contains infinitely many $x_n$. Take $r=1$ and see that this is impossible, as $d(x_n, x_m) < 2$ infinitely many times...
We do not need limit point compactness implies compactness. (this exercise could be seen as a step toward the proof of this fact).
1 implies 2: Indeed if $\phi: X \to \mathbb{R}$ is continuous then $F(x) = (x,\phi(x)): X \to X \times \mathbb{R}$ is an embedding: the family $\{1_X, \phi\}$ separates points and points and closed sets, using a standard theorem (34.2 in Munkres).
This means that the product metric restriced to $F[X]$ i.e. 
$$d_\phi(x,x') = d(x,x') + |\phi(x) - \phi(x')|$$ is an equivalent metric for $(X,d)$ as well. (this can also be shown directly). As $d_\phi$ is bounded by 1., so must $\phi$ be. So 2. follows
