Compactness is a topological property, proof without open covers An exercise in the book wants me to show that compactness is a topological property. This chapter is about topology of $R^n$ and we are working with an assumed metric. We haven't come to definition of topological spaces yet. We defined; open sets, closed sets, limit points, sequences, continuous functions, bounded sets, homeomorphisms and topological properties etc.
Here is the definition of compactness that the book uses.
Defn: A set $S$ is compact if every infinite sequence contained in $S$ has a limit point contained in $S$.
Now I am trying to prove the following.

Proposition: Compactness is a topological property.
I am trying to show that if $A \subset R^n$ is compact and there exists a homeomorphism $f: A \to B$, then $B$ is compact. 
Strategy: Assume $A$ is compact, and $f$ is continuous with a continuous inverse. Go with proof by contradiction. Suppose $B$ is not compact. Then by Heine-Borel, $B$ is not (closed and bounded). So $B$ is either not closed or not bounded. 
I showed that if $B$ is not closed, then $A$ is not compact and found a contradiction. I am trying to also show that if $B$ is not bounded, a contradiction arises. Then I think, I can conclude $B$ must be compact because all roads lead to a contradiction. 
I need to show if $B$ is not bounded, a contradiction arises. Possible candidate is to show that $A$ is not compact, which contradicts the hypothesis.
 A: If $A$ is closed then $B$ is closed because a homeomorphism is a closed map. Now we prove that $B$ is also bounded, by exploiting the fact that a subset of $\mathbb{R}^n$ is compact if and only if is sequentially compact.
So, assume by contradiction that $B$ is not bounded. Then we can find a sequence $\{b_n \} \subset B$ such that $||b_n|| > n$ for all $n \in \mathbb{N}$. This sequence clearly does not admit any convergent subsequence. 
On the other hand, we have $b_n = f(a_n)$ with $a_n \in A$, and by compactness of $A$ we can find a subsequence $\{a_{n_k}\}$ of $\{a_n\}$ converging to $a \in A$. So, by continuity of $f$, the subsequence $\{b_{n_k}\} = \{f(a_{n_k})\}$ should converge to $f(a)$, contradiction.
Then $B$ is also bounded, hence compact by Heine-Borel.  
A: Let $(y_n)$ be an infinite sequence in $B.$ As $f$ is surjective, $$y_n=f(x_n)$$ where $(x_n)$ is s sequence in $A.$ Injectivity implies that $(x_n)$ is also an infinite sequence. Since, $A$ is compact, therefore $(x_n)$ has a convergent subsequence (denoted again by $(x_n)$, for convenience) in $A,$ to say $x$.
Thus, we have $$x_n \to x\text{ as } n\to \infty$$
Continuity of $f$ implies $$y_n=f(x_n) \to f(x)\text{ as } n\to \infty$$
Thus, $f(x)\in B$ is a limit point of $(y_n).$
A: You have to use open covers (for all spaces not just $\mathbb{R}^n$): a homeomorphism need not preserve boundedness at all.
A: The definition of compactness you have is what is in general topological spaces known as sequential compactness. It turns out that it is equivalent to compactness defined by open covers in metric spaces.
Now, the proof should use definition directly. If you have $f\colon A\to B$ a homeomorphism, and assume that $A$ is compact, then take any sequence $(b_n)$ in $B$. Sequence $(f^{-1}(b_n))$ is sequence in $A$, so it has convergent subsequence $(f^{-1}(b_{p(n)}))$ with limit $a$. Since $f$ is continuous, we have that $$f(a) = \lim_n f(f^{-1}(b_{p(n)})) = \lim_n b_{p(n)},$$ so $(b_n)$ has convergent subsequence.
