Forward direction $ (\Longrightarrow) $
Suppose that $ T \in \mathcal{L}(X,Y) $ and that $ x_{n} \stackrel{\text{wk}}{\longrightarrow} x $. Then for any $ \varphi \in Y^{*} $, we have
\begin{align}
\lim_{n \to \infty} \varphi(T(x_{n}) - T(x))
&= \lim_{n \to \infty} \varphi(T(x_{n} - x)) \\
&= \lim_{n \to \infty} (\varphi \circ T)(x_{n} - x) \\
&= 0. \quad (\text{As $ \varphi \circ T \in X^{*} $.})
\end{align}
Therefore, $ T(x_{n}) \stackrel{\text{wk}}{\longrightarrow} T(x) $.
Backward direction $ (\Longleftarrow) $
Suppose that $ T: X \to Y $ is a linear operator (not assumed to be bounded) and that $ x_{n} \stackrel{\text{wk}}{\longrightarrow} x $ implies $ T(x_{n}) \stackrel{\text{wk}}{\longrightarrow} T(x) $.
For the sake of contradiction, assume that $ T $ is not bounded. Then we can easily construct a sequence $ (x_{n})_{n \in \mathbb{N}} $ in $ X $ that converges in norm to $ 0_{X} $ but for which $ \| T(x_{n}) \|_{Y} \geq n^{2} $ for all $ n \in \mathbb{N} $. As $ \dfrac{1}{n} \cdot x_{n} \stackrel{\| \cdot \|_{X}}{\longrightarrow} 0_{X} $, we obviously have $ \dfrac{1}{n} \cdot x_{n} \stackrel{\text{wk}}{\longrightarrow} 0_{X} $. From our initial hypothesis, it follows that $ T \left( \dfrac{1}{n} \cdot x_{n} \right) \stackrel{\text{wk}}{\longrightarrow} T(0_{X}) = 0_{Y} $.
Now, for each $ n \in \mathbb{N} $, we can view $ T \left( \dfrac{1}{n} \cdot x_{n} \right) $ as an element $ \Psi_{n} $ of $ Y^{**} $, via the canonical embedding $ J: Y \to Y^{**} $. As
\begin{align}
\forall \varphi \in Y^{*}: \quad
\lim_{n \to \infty} {\Psi_{n}}(\varphi)
&= \lim_{n \to \infty} \varphi \left( T \left( \frac{1}{n} \cdot x_{n} \right) \right) \\
&= \varphi(0_{Y}) \\
&= 0,
\end{align}
we see that $ \displaystyle \sup_{n \in \mathbb{N}} |{\Psi_{n}}(\varphi)| < \infty $ for all $ \varphi \in Y^{*} $. Applying the Uniform Boundedness Principle (also known as the Banach-Steinhaus Theorem) to $ Y^{**} $, we obtain $ \displaystyle \sup_{n \in \mathbb{N}} \| \Psi_{n} \|_{Y^{**}} < \infty $. This yields $ \displaystyle \sup_{n \in \mathbb{N}} \left\| T \left( \dfrac{1}{n} \cdot x_{n} \right) \right\|_{Y} < \infty $ as the canonical embedding $ J $ is an isometry. We have thus arrived at a contradiction as our construction of the sequence $ (x_{n})_{n \in \mathbb{N}} $ forces us to have $ \left\| T \left( \dfrac{1}{n} \cdot x_{n} \right) \right\|_{Y} \geq n $ for all $ n \in \mathbb{N} $ instead.
The assumption is therefore false, so we conclude that $ T $ is indeed bounded.
Note: There is no need to assume that $ X $ is reflexive.
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 $T$.
Best Answer
The reverse implication is a simple consequence of the closed-graph theorem: Let $(x_n)$ be a sequence such that $x_n\to x$ and $Tx_n\to y$. We need to show $Tx=y$. Since every strongly convergent sequence is weakly convergent, it follows $x_n\rightharpoonup x$, and by the assumption $Tx_n \rightharpoonup Tx$. Since weak limits are unique, $Tx=y$, the graph of $T$ is closed, and $T$ is continuous.