I came across this question:
In a compact metric space $(X,d)$ if every convergent subsequence of a sequence converges to the same limit, say $l$, then the original sequence also converges to $l$.
Sequences and Series – Convergence of Sequence with Convergent Subsequences
compactnessmetric-spacessequences-and-series
Related Solutions
I think that the argument in the linked post should filter through fine (in fact, it seems constructed for general reflexive spaces).
A reflexive Banach space is separable if and only if its dual is. (The standard result is that $X^*$ separable implies $X$ separable.) Fix a countable dense subsequence $\{\phi_1,\phi_2,\ldots\}\subseteq X^*$, and let $\{a_1,a_2,\ldots\}\subseteq A$ be any sequence. Let them all be bounded by $M$ in norm, since $A$ was bounded. Consider $\{\phi_1(a_1),\phi_2(a_2),\ldots\}\subseteq \mathbb{K}$. This is a bounded sequence in a locally compact space (bounded because all of them have norm less than or equal to $\|\phi_1\|\cdot M$), so it has a convergent subsequence $\phi_1(a_{11}),\phi_2(a_{12}),\ldots$. Now for each $k+1$, similarly extract a convergent subsequence from $\phi_{k+1}(a_{1k}),\phi_{k_1}(a_{2k}),\ldots$. Do a standard diagonal argument and $a_{11},a_{22},\ldots$ is a sequence for which each functional $\phi_i$ converges. Moreover, because the $\phi_i$ were dense, for each $\phi\in X^*$, $\phi(a_{11}),\phi(a_{22}),\ldots$ is convergent. Define the functional $a$ on $X^*$ by $a(\phi)=\lim\limits_{n\to\infty} \phi(a_{nn})$. This is continuous (needs checking) and therefore by reflexivity is represented by some element $x\in X$. Then the sequence $\{a_{nn}\}\xrightarrow{w} x$, and because $A$ was $w$-closed, $x$ is in $A$.
For general spaces, fix your sequence $\{a_1,\ldots\}$ and let $Y$ be its closed linear span. Then $Y$ is separable, and it is a closed subspace of a reflexive space, so itself reflexive (A closed subspace of a reflexive Banach space is reflexive) and the dual is separable. But now we're in the old case, so we extract $x$ with $y^*(a_n)\rightarrow y^*(x)$ for each $y^*\in Y^*$. Now this convergence holds for any linear functional in $X^*$, because the restriction to $Y$ is a linear functional on $Y$.
This is true but not very easy to prove.
Suppose $X$ is not compact. Without loss of generality assume that the original metric $d$ on $X$ is such that $d(x,y)<1$ for all $x,y\in X.$ There exists a decreasing sequence of non-empty closed sets $\{C_{n}\}$ whose intersection is empty. Let $$\rho (x,y)=\sum_{n=1}^{\infty }\frac{1}{% 2^{n}}d_{n}(x,y)$$ where $$d_{n}(x,y)=\left\vert d(x,C_{n})-d(y,C_{n})\right\vert +\min \{d(x,C_{n}),d(y,C_{n})\}d(x,y).$$ We claim that $\rho $ is a metric on $X$ which is equivalent to $d$ and that $% (X,\rho )$ is not complete. Note that $d_{n}(x,y)\leq 2$ for all $x,y\in X.$ If $x$ and $y$ $\in C_{k}$ then $x$ and $y$ $\in C_{n}$ for $1\leq n\leq k$ and hence $\rho (x,y)\leq \sum_{n=k+1}^{\infty }\frac{2}{2^{n}}=% \frac{1}{2^{k}}$. Thus, the diameter of $C_{k}$ in $(X,\rho )$ does not exceed $\frac{1}{2^{k}}$. Once we prove that $\rho $ is a metric equivalent to $d$ it follows that $\rho $ is not complete because $\{C_{n}\}$ is a decreasing sequence of non-empty closed sets whose intersection is empty.
Assuming (for the time being) that $d_{n}$ satisfies triangle inequality it follows easily that $\rho $ is a metric: if $\rho (x,y)=0$ then $% d(x,C_{n})=d(y,C_{n})$ for each $n$ and $\min \{d(x,C_{n}),d(y,C_{n})\}d(x,y)=0$ for each $n$. If $d(x,y)\neq 0$ it follows that $d(x,C_{n})=d(y,C_{n})=0$ for each $n$ which implies that $x$ and $y$ belong to each $C_{n}$ contradicting the hypothesis. Thus $\rho $\ is a metric. Also $\rho (x_{j},x)\rightarrow 0$ as $j\rightarrow \infty $ implies $\left\vert d(x_{j},C_{n})-d(x,C_{n})\right\vert \rightarrow 0$ and $% \min \{d(x_{j},C_{n}),d(x,C_{n})\}d(x_{j},x)\rightarrow 0$ as $j\rightarrow \infty $ for each $n$. There is at least one integer $k$ such that $x\notin C_{k}$ and we conclude that $d(x_{j},x)\rightarrow 0$. Conversely, suppose $% d(x_{j},x)\rightarrow 0$. Then $d_{n}(x_{j},x)\rightarrow 0$ for each $n$ and the series defining $\rho $ is uniformly convergent, so $\rho (x_{j},x)\rightarrow 0$. It remains only to show that $d_{n}$ satisfies triangle inequality for each $n$ . We have to show that $$\left\vert d(x,C_{n})-d(y,C_{n})\right\vert$$ $$+\min \{d(x,C_{n}),d(y,C_{n})\}d(x,y)$$
$$\leq \left\vert d(x,C_{n})-d(z,C_{n})\right\vert +\min \{d(x,C_{n}),d(z,C_{n})\}d(x,z)$$ $$+\left\vert d(z,C_{n})-d(y,C_{n})\right\vert +\min \{d(z,C_{n}),d(y,C_{n})\}d(z,y)$$ for all $x,y,z.$ Let $% r_{1}=d(x,C_{n}),r_{2}=d(y,C_{n}),r_{3}=d(z,C_{n})$. We consider six cases depending on the way the numbers $r_{1},r_{2},r_{3}$ are ordered. It turns out that the proof is easy when $r_{1}$ or $r_{2}$ is the smallest of the three. We give the proof for the case $r_{3}\leq r_{1}\leq r_{2}$. (The case $r_{3}\leq r_{2}\leq r_{1}$ is similar). We have to show that
$$r_{2}-r_{1}+r_{1}d(x,y)\leq r_{1}-r_{3}+r_{3}d(x,z)+r_{2}-r_{3}+r_{3}d(z,y)$$ which says $$r_{1}d(x,y)\leq 2r_{1}-2r_{3}+r_{3}d(x,z)+r_{3}d(z,y).$$ Since $d$ satisfies trangle inequality it suffices to show that $$% r_{1}d(x,z)+r_{1}d(z,y)\leq 2r_{1}-2r_{3}+r_{3}d(x,z)+r_{3}d(z,y).$$ But this last inequality is equivalent to $$(r_{1}-r_{3})[d(x,z)+d(z,y)]\leq 2r_{1}-2r_{3}.$$ This is true because $d(x,z)+d(z,y)\leq 1+1=2$.
Best Answer
Suppose the sequence does not converge to $l$. Then $\exists$ an $\epsilon > 0$ such that for every $n_{0} \in \mathbb{N}, \exists n > n_{0}$ such that $d(x_{n},l) \geq \epsilon$. Thus, we get a subsequence $x_{n_{k}}$ such that $d(x_{n_{k}},l) \geq \epsilon$ for every $k \in \mathbb{N}$. Now using the compactness of the space, this subsequence has a convergent subsequence, say $x_{n_{k_{m}}}$, which cannot converge to $l$, however. But this subsequence is also a subsequence of the original subsequence and hence should converge to $l$. Hence the contradiction.