A metric space is compact iff it is pseudocompact I was recently presented this problem from a course on topology half of which I could work out but the other half is a mystery:

Let $ (X, \tau) $ be a metrizable topological space, we say that a metric $ d:X \times X \to R $ is continuous if it is a continuous function with the product topology on $ X \times X $ and the standard Euclidean topology on $ R $.
a. We are to quote the Tychonoff theorem (product of compacts is again compact).
b. We are to show that if X is compact then every continuous metric d is bounded. This I could do as simply given X is compact so is $ X \times X $ by Tychonoff theorem and if d is a continuous mapping then we know that $ d:X \times X \to R $ is a continuous function from compact set is its image is compact so by Heine-Borel theorem its range is bounded which is the proof.
c. We are to show that if every continuous metric on $ X \times X $ is bounded then every function in $ C(X,R) $ is bounded. I could not solve this unfortunately.
d. We are to show equivalence of thefollowing three statements: (Could not solve except i->ii due to previous parts)
i. X is compact
ii. Every function in C(X,R) is bounded
iii. (X,d) is bounded under any continuous metric d

As you can see my problems lie in part c and parts of part d so I need the help on those as I have tried to think of a solution but nothing came up. Help needed and kindly appreciated.
 A: It seems the following.
c. Let $d$ be a continuous metric on $ X \times X $. If $f\in C(X,\Bbb R)$ in an unbounded function, then $d’(x,y)=d(x,y)+|f(x)-f(y)|$ is an unbounded continuous metric on $ X \times X $.
d. ii $\Rightarrow$ iii. Assume that $d$ is an unbounded continuous metric on $ X \times X $. Fix an arbitrary point $x_0\in X$ and put $f(x)=d(x_0,x)$. The unboundedness of the metric $d$ and triangle inequality imply that the continuous function $f$ is unbounded too.  
iii $\Rightarrow$ ii. It is Claim c.
ii $\Rightarrow$ i. 
If $X$ is a Tychonoff space and each continuous real-valued function on the space $X$ is bounded, then the space $X$ is called pseudocompact. A topological space $X$ is countably compact if each infinite subset of $X$ has a cluster point. Each compact space is countably compact and each countably compact space is pseudocompact, and both inclusions are strict. But for metric spaces these conditions coincide. It is well-known because each normal pseudocompact space is countably compact [Eng] and each countably compact regular space with $G_\delta$-diagonal is a metrizable compact [Gru]. 
But we can give a direct proof. By Theorems 4.3.27-29 from [Eng], a metric space $(X,d)$ space is compact iff $(X,d)$ is complete and totally bounded. 
If the space $(X,d)$ is not complete then let $(X’, d’)$ be a completion of the space $(X,d)$. Since the space $(X,d)$ is incomplete, there exists a point $x_0\in X'\setminus X$. Consider a function $f$ such that $f(x)=1/d’(x,x_0)$ for all $x\in X$. Since  $d’$ is a extension of the metric $d$, and $x_0\not \in X$, the function $f$ is continuous. The density of the set $X$ in the space $X'$ imply that the function $f$ is unbounded.
If the space $(X,d)$ is not totally bounded, then there exist a number $\varepsilon>0$ and a infinite subset $X_0=\{x_n\}$ of the space $X$ such that $d(x_m, x_n)\ge 3\varepsilon$ for each pair of distinct indices $m$ and $n$. Then $X_0$ is a closed discrete subset of the space $X$ so a function $f_0:X_0\to\Bbb R$, $f_0(x_n)=n\varepsilon$ is continuous. We can extend the function $f_0$ to a continuous function $f$ from $X$ to $\Bbb R$ by Tietze-Urysohn theorem, of directly, by putting for each $x\in X$
$$f(x)=\sum_{n=1}^\infty \max\{0, n(\varepsilon - d(x,x_n)) \}. $$
References
[Eng]  Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.



[Gru] Gary Gruenhage Generalized Metric Spaces, in: K.Kunen, J.E.Vaughan (eds.) Handbook of Set-theoretic
Topology, Elsevier Science Publishers B.V., 1984.


