Proving if a function is continuous, its inverse is also continuous 
Let $(X,d)$ and $(X,d')$ be metric spaces. Prove that $a)\implies b) $ and $b) \implies c)$:
a) if $x_n \to x$ in $(X,d')$ then $x_n \to x$ in $(X,d)$
b)For any metric space $(Y,p)$ and any continuous $f:(X,d)\to (Y,p), f:(X,d')\to(Y,p)$ is also continuous.
c) for any metric space $(Y,p)$ and any continuous $f:(Y,p)\to(X,d'), f:(Y,p)\to(X,d)$ is also continuous.

$f: E\to E'$ a continuous function. Prove that if $E$ is compact and $f$ is bijective then $f^{-1}:E' \to E$ is continuous.
I tried to prove $a)\implies b)$ by if $d'(x',x)<\delta$ then there exists some sequence $\{x_n\}$ such that $x_n \to x$ in $(X,d')$. Then $x_n \to x$ in $(X,d)$ and there exsits a $N$ such that $d(x_n,x)<\delta$ whenever $n>=N$. Then $d(x',x)<\delta$ for all $x\in \{x_n|n>=N\}$.   (I'm not sure if my proof is valid)
For $b) \implies c)$, I wanted to prove if $f$ is continuous then so is $f^{-1}$. But I'm not sure if it's generally true.

Hint:
If b) is true then the function $(X,d')\to (X,d)$ prescribed by $x\mapsto x$ is continuous. 
Let's denote this function with $g$. 
Now if $f:(Y,p)\to(X,d')$ is continuous then so is $g\circ f:(Y,p)\to(X,d)$.
