If $X$ is compact then $X$ is sequentially compact. My attempt :
We assume that $X$ is a compact metric space and we let $A=(x_n)$ be a sequence. Then $A \subset G_1 \cup G_2.....G_k$.So suppose $G_i$ contains infinitely many points of $(x_n)$.We choose $y \in G_i$ in such a way so that  we can find a ball $B_d(y,\epsilon) $ such that it is contained in $G_i$ and it contains infinitely many points of $(x_n)$.Then let us say that the susbsequence $(x_{n_k})$ is contained in the ball. I am trying to show that the subsequence $(x_{n_k})$ converges to $y$.
How do I go about from here? If someone could suggest a way out instead of suggesting a different answer.
 A: One possible approach: suppose $(x_n)$ has no convergent subsequence. Then, for each integer $n$ there is an open neighborhood $U_i\ni x_i$ such that $U_i\cap U_j=\emptyset$ for all $i\neq j.$ Furthermore, $C:=\{x_n\}_{n\in \mathbb N}$ is closed because it has no limit points, hence has them all. Then $\mathscr U=\{C^c\}\cup \{U_i\}_{i\in \mathbb N}$ is an open cover with no finite subcover.
A: You need to state explicitly that $X$ is a metric space, since the result is not true for topological spaces in general. The beginning of your argument makes no sense as stated: where do $G_1,\ldots,G_k$ come from, and what are they? We might guess that you started with some open cover of $X$ and took a finite subcover, but you need to say so. You also need to say something about the open cover with which you started. For all we know, you started with the cover $\{X\}$, so that $k=1$ and $G_1=X$.
Next, for a given $\epsilon$ how do you know that there is point $y\in G_i$ such that $B_d(y,\epsilon)\subseteq G_i$ and $B_d(y,\epsilon)$ contains infinitely terms of the sequence? This requires proof, which you have not given. Finally, it is simply not true that the subsequence in question necessarily converges to $y$.
At this point you don’t have enough of a proof for it to be salvageable. I strongly recommend that you try instead to prove the result by showing that if $X$ has a sequence with no convergent subsequence, then $X$ is not compact. A direct proof is possible, by showing that every sequence in $X$ has a Cauchy subsequence and then using the fact that compact metric spaces are complete, but that’s a bit harder.
A: Suppose $X$ is not sequentially compact, then there is a sequence $(p_n)\subset X$ which admits no subsequence converging to a point in $X$. Then $\forall x\in X,\exists r\gt 0$ such that $p_j\in B_d(x,r)$ for only finitely many $j$. $\{{B_d(x,r)}:x\in X\}$ is an open cover of $X$, so by compactness it admits a finite subcover $\{B_d(x_i,r_i)\}_{i=1}^N$. Since $(p_n)$ have only finitely many terms in each $B_d(x_i,r_i)$, it follows that $(p_n)$ has only finitely many terms, a contradiction.
