Given a continuous function $f(x):[a,b] \to[c,d] $ and its inverse $g(x) : [c,d]\to [a,b]$ prove the continuity of $g(x)$
Since $f$ is continuous I thought I can write for $x=x_1$,$\lim_{x\to x_1} = f(x_1)$ due to continuity, and, $|x-x_1| \lt \delta \implies |f(x)-f(x_1)| \lt \varepsilon$.
But now for $g(x)$ I get $|x-x_1| = |g(f(x)) - g(f(x_1))| \lt \delta \implies |f(x)-f(x_1)| \lt \varepsilon$
But rather I understand what I need is $|f(x)-f(x_1)| \lt \delta \implies |g(f(x))-g(f(x_1))| \lt \varepsilon$ which I am not able to do
Also is it justified to take for granted $g(f(x)) = x$ for all $x\in[a,b]$ ?
Yes, we have $g(f(x)) = x$ for all $x \in [a,b]$. Furthermore $f(g(y))=y$ for all $y \in [c,d].$
With sequences: let $y_0 \in [c,d]$ and $(y_n))$ a sequence in $[c,d]$ such that $y_n \to y_0$. We have to show that $g(y_n) \to g(y_0)$.
To this end put $x_n:=g(y_n)$. Then $(x_n)$ is a sequence in $[a,b]$. Let $x_0$ be an accumulation point of $(x_n)$. Then there is a subsequence $(x_{n_k})$ such that $x_{n_k} \to x_0$ as $k \to \infty$.
Since $f$ is continuous, we have $y_{n_k}=f(x_{n_k}) \to f(x_0)$.
This gives $f(x_0)=y_0$, hence $x_0=g(y_0)$.
Conclusion: the bounded sequence $(g(y_n))$ has exactly one accumulation point: $g(y_0)$.
This shows that $g(y_n) \to g(y_0)$.