Understanding the last part in the proof that $T$ is bounded. I am trying to understand the proof of the reverse direction in the first answer in this link:
Prove that $T$ is bounded iff $x_{n} \rightharpoonup x \quad \Rightarrow \quad Tx_{n} \rightharpoonup Tx.$
Here is the proof I am speaking about:
Using reflexivity of $X$:
For the converse, you need to use that $X$ is reflexive. We want to show that $T$ is continuous: that is, if $x_n\to x$, then $Tx_n\to Tx$. Because of linearity, it is enough to show that $x_n\to0$ implies $Tx_n\to0$. So suppose that $x_n\to0$. Then of course $x_n\rightharpoonup0$, which by hypothesis implies that $Tx_n\rightharpoonup0$. Because every weakly convergence sequence is bounded, the sequence $\{Tx_n\}$ is bounded. Because $X$ is reflexive, closed balls are weakly compact. So there exists a convergent subsequence $\{Tx_{n_k}\}$. Say $Tx_{n_k}\to y$. Since strong convergence implies weak convergence, $y=0$. Now we can apply this last reasoning to every sequence of $\{Tx_n\}$: that is, any subsequence of $\{Tx_n\}$ has a subsequence that converges to $0$: so $Tx_n\to0$. Thus $T$ is continuous, so bounded. 
My questions are:
1-Is there is a typo in this part of the second line from below"apply this last reasoning to every sequence of $\{Tx_n\}$," Should it be every subsequence instead?
2-Also I do not know how to prove this lemma:
Lemma Let $X$ be a topological space and $\mathbf{x}=(x_n)_{n\in \mathbb{N}}$ be a sequence of elements of $X$. If every subsequence of $\mathbf{x}$ contains a subsequence convergent to $x$ then $x_n \to x$.
Could anyone provide me with the proof please?  
3- I do not understand the finalization of the proof in the last 2 lines, could anyone explain it for me?

To answer your questions: 
1) You are correct, it should say subsequence.  
2) You can prove this by contradiction. We prove it for the case of metric spaces (since you are in a normed vector space this is enough). Assume that $x_n \not\to x$. Then there exists $\epsilon > 0$ and an increasing sequence $(n_k)_{k \geq 1}$ such that $d(x, x_{n_k}) \geq \epsilon$ for all $k \geq 1$. This subsequence cannot have a further subsequence converging to $x$ leading us to a contradiction. 
3) I'm not sure what you are asking. The proof shows that for any $x_n \to 0$ we have $Tx_n \to 0$. By linearity this shows that $x_n \to x$ implies $Tx_n \to Tx$. This is equivalent to continuity of $T$ which in turn is equivalent to boundedness of $T$.
