Proof that continuous function has continuous inverse Assume that $A\subseteq\mathbb{R}^n$ is compact and that $\mathbf{f}:A\to\mathbb{R}^m$ is continuous and injective. Let $B=\mathbf{f}(A)\subseteq\mathbb{R}^m$ and show that $\mathbf{f}^{-1}(B)\rightarrow A$ is continuous. 
I want to use the sequential definition of continuity.
Let $\mathbf{y} \in B$ and $(\mathbf{y}_n) \rightarrow \mathbf{y}$ with $\mathbf{y}_n \in B$ for each $n\in \mathbb{N}$. 
Then since $B=\mathbf{f}(A)$, there are $\mathbf{x} , \mathbf{x}_n \in A$ such that  $\mathbf{f}(\mathbf{x}) =\mathbf{y}$ and $\mathbf{f}(\mathbf{x}_n) = \mathbf{y}_n$ for each $n$. 
Since $A$ is compact, it is closed and bounded. So $(\mathbf{x}_n)$ is bounded.  We show that $(\mathbf{x}_n)$ converges.
It's enough to show that any convergent subsequence of $(\mathbf{x}_n)$ converges to the same point in $\mathbb{R}^n$. So let $(\mathbf{x}_{n_k})$ be a convergent subsequence that converges to $\mathbf{a}$. Since $A$ is closed, $\mathbf{a} \in A$. By continuity, $\mathbf{f}(\mathbf{x}_{n_k})\rightarrow \mathbf{f}(\mathbf{a})$. Therefore, $(\mathbf{y}_{n_k})\rightarrow \mathbf{f}(\mathbf{a})$. But since $(\mathbf{y}_{n_k})$ is a subsequence of $(\mathbf{y}_n)$, we get that $\mathbf{y} =\mathbf{f}(\mathbf{a})$. Thus, $\mathbf{f}^{-1}(\mathbf{y})=\mathbf{a} =\lim_{k\to \infty} \mathbf{x}_{n_k}$. And every convergent subsequence of $ (\mathbf{x}_{n})$ converges to $\mathbf{f}^{-1}(\mathbf{y}):=\mathbf{a}$, so that $(\mathbf{x}_n)\rightarrow \mathbf{a}$. 
EDITED QUESTION: Does this proof work? I'm a bit concerned that I haven't used the fact that $\mathbf{f}({A})$ is a compact set. Overall, I'm not sure if I made a mistake somewhere. 
 A: Start from your sequence $(\mathbf{y_n})\to \mathbf{y}$ and consider an arbitrary subsequence $(\mathbf{y_{n_k}})$ of $(\mathbf{y_n})$. By what you proved, you can find a subsequence $(\mathbf{y_{n_{k_j}}})$ of $(\mathbf{y_{n_k}})$ such that $(f^{-1}(\mathbf{y_{n_{k_j}}}))\to f^{-1}(\mathbf{y})$. This implies that the entire sequence $(f^{-1}(\mathbf{y_{n}}))\to f^{-1}(\mathbf{y})$ since otherwise, you could find a subsequence  $(\mathbf{y_{n_k}})$ such that $\vert f^{-1}(\mathbf{y_{n_{k}}})-f^{-1}(\mathbf{y})\vert\ge \varepsilon>0$ for all $k$, which would be a contradiction.
I am using this fact
theorem on subsequences
Does this answer your question?
A: *

*Since $f$ is continuous, $f(A)$ is compact and therefore closed.

*Let $y_{n}\in f(A)$ a convergent sequence. By 1., we can supose $y_{n}=f(x_{n})\to f(x)=y$.

*We want to show that $f^{-1}(y_{n})\to f^{-1}(y)$, i.e. $x_{n}\to x$. If $x_{n}\not\to x$, then there exists a subsequence such that $|x_{n_{k}}-x|\geq \epsilon$. Then we can extract a convergent subsequence $x_{n_{k_{j}}}$ of $x_{n_{k}}$ to, say, $x'$. But then we would have $f(x')\leftarrow f(x_{n_{k_{j}}})=y_{n_{k_{j}}}\to y=f(x)$. So, $f(x)=f(x')$ and then $x=x'$, a contradiction.
