Your part 2 is slightly wrong. We need to show that $(x_n)$ $\in A$ Cauchy implies it converges to a limit in $A$. (This is different to what you have said, see my next paragraph) We do this by saying $X$ is complete, so $(x_n) \rightarrow x \in X$ and then use the fact that $A$ is closed in $X$ to say that since $(x_n) \rightarrow x \in X$, that $x\in A$.
This is not what you have said, since you mention that every convergent sequence is Cauchy (true) but it is not true in general that every Cauchy sequence is a convergent sequence (i.e. in incomplete metric spaces)
The fact that complete and closed are different concepts is because you can have closed sets in incomplete metric spaces. However, this theorem shows us that in any complete metric space, completeness and closure of a set is the same thing.
I already showed in this answer that $X'$ relatively compact implies $X'$ relatively countably compact. Also, if $X'$ is relatively compact, the $\overline{X'}$ is compact, and thus sequentially compact (this holds in particular in metric spaces, but also more broadly). So any sequence from $X'$ has a convergent subsequence with limit in $\overline{X'}$, so $X'$ is then relatively sequentially compact as well.
In any space, $X'$ relatively sequentially compact implies $X'$ relatively countably compact: any infinite subset $A$ of $X'$ contains some sequence with all different elements, which has a convergent subsequence to some $x \in X$, and this $x$ is an accumulation point of $A$.
If $X'$ is relatively countably compact, and $X$ is metric (first countable and $T_1$ will already do), let $(x_n)$ be a sequence from $X'$. If $A = \{x_n: n \in \mathbb{N}\}$ is finite, some value occurs an infinite number of times, and yields a convergent subsequence. So assume $A$ is infinite, so it has an accumulation point $p \in X$. Because $X$ is $T_1$, this means that every neighbourhood of $p$ intersects $A$ in infinitely many points. Pick $x_{n_1}$ in $B(p, 1)$, $x_{n_2}$ with $n_2 > n_1$ in $B(p, \frac{1}{2})$, and so on, by recursion. This defines a convergent subsequence of $(x_n)$ that converges to $p$. So $X'$ is relatively countably compact.
If $X'$ is relatively countably compact, and $X$ is metric, then we do get that $X'$ is relatively compact (this is due to Hausdorff, IIRC). I cannot reconstruct a proof right away, but there is one here, e.g. (but this involves Cauchy filters etc.)
It turns out that these notions are also equivalent in some topological vector spaces (spaces of the form $C_p(X)$ where $X$ is compact (Grothendieck), and weak topologicals on normed vector spaces (Eberlein-Smulian)), but these are a bit more involved, I think.
Best Answer
If $v_n$ is a Cauchy sequence, $v_n$ is bounded. To see this suppose $\varepsilon = 1$ is given. Then there is $N\in \mathbb{N}$ such that $||v_n- v_m||< 1$ for $n, m\ge N. $ In particular, $$||v_n|| = ||v_n - v_N + v_N||\le ||v_N|| + 1$$ for $n\ge N$. Since there are only finitely many $k< N$ the sequence as a whole is bounded.
So $(v_n)$ is contained in some closed ball, which (due to your assumptoins about sequential compactness) implies it has a convergent subsequence $v_{n_k}$ with $$\lim_k v_{n_k} =: v$$ say. Since $v_n$ is Cauchy, it is now not difficult to see that it is actually the whole sequence $(v_n)$ which converges to $v$ - this I leave as a final exercise to you.