I'm trying to prove the following result:
Given a set $A$ in $\mathbb{R}^n$ with the usual topology given by the euclidean distance. If all infinite subset $B \subset A$ has an accumulation point in $A$, then $A$ in sequentially compact.
This is a partial result of the Heine-Borel-Lebesgue's Theorem.
Please, could you give me a sketch of the proof.
Thank you very much.
As hinted by @Danny Duberstein in the comments, a good candidate for the limit of a convergent subsequence of $(a_i)$ is the limit (accumulation) point $b$ of $B = \{ a_i \}$, where $b \in A$ by hypothesis.
To show that there is indeed a subsequence $(a_{i_j}) \to b$, consider, for each $j$, the basis element $U_j = A \cap B_2 (b, \frac{1}{j})$. Since $b$ is a limit point of $B$, choose some point of the sequence $a_{i_j} \in B \cap U_j$. The corresponding subsequence $(a_{i_j})$ can be seen to converge to $b$. Any neighborhood $V$ of $b$ in $A$ must contain some basis element $A \cap B_2 (b, \varepsilon)$. Take $J = \lfloor \frac{1}{\varepsilon} \rfloor$; note that $\frac{1}{j} < \varepsilon$ and therefore $U_n = A \cap B_2 (b, \frac{1}{j}) \subseteq A \cap B_2 (b, \varepsilon) \subseteq V$ for each $j > J$. Since $a_{i_j} \in U_j$ for each $j$, it follows that $a_{i_j} \in V$ for each $j > J$.