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.

Your proof for (a) $\Rightarrow$ (b) is not valid, because i could not find $f$ anywhere in your proof.
Hint:
It suffices to use only sequences, no need for $\varepsilon,\delta, n\geq N$ stuff!
Note that a function in metric spaces is continuous iff for all convergent seqences $x_n\to x$
we have the convergence $f(x_n)\to f(x)$.
Hint for (b) $\Rightarrow$ (c):
Apply (b) for $(Y,p)=(X,d)$ and $f$ as the identity, then
use the composition of two continuous functions.
