I have to prove that if $T:(E,\|\cdot\|_E)\rightarrow (F,\|\cdot\|_F)$ is a continuous and linear operator, and $x_h\rightharpoonup x$ in $E$, than $Tx_h\rightharpoonup Tx$ in $F$. So we know that $T$ is continuous with respect to the strong topologies, and we want to prove that it is also continuous with respect to the weak ones.
Is there a simple proof of this fact?
I've just noticed that I didn't understand your question at the first time. So I'll change my answer almost completely.
I'll try to answer it in a general setting. Given a set $X$, a family $\{(X_i,\tau_i):i\in I\}$ of topological spaces and a famlily of functions $\mathcal{F}=\{f_i\colon X\to X_i|i\in I\}$, let $\sigma(X,\mathcal{F})$ be the least topology on $X$ such that all the functions of $\mathcal{F}$ are continuous. This topology is generated by finite intersections of sets of the form $f_i^{-1}(V)$, for $i\in I$ and $V\in \tau_i$.
Notice that the weak topology is a particular case of this type of topology.
In this context, we have the following
Fact 1. Let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $X$ and let $x\in X$. Then $x_n\to x$ in $\sigma(X,\mathcal{F})$ if and only if $f_i(x_n)\to f_i(x)$ in $\tau_i$ for all $i\in I$.
The proof is hidden below:
Suppose that $x_n\to x$ in $\sigma(X,\mathcal{F})$. If $f_i(x)\in V\in\tau_i$, then $x\in f_i^{-1}(V)$, hence $x_n\in f_i^{-1}(V)$ for $n\gg 0$, from which it follows that $f_i(x_n)\in V$ for $n\gg 0$.
${}$
Conversely, if $U\in\sigma(X,\mathcal{F})$, then $U=\bigcap_{j\leq n}f_{i_j}^{-1}(U_j)$ for $i_0,\dotso,i_n\in I$ and $U_{i_j}\in\tau_{i_j}$ for each $j\leq n$. If $x\in U$, then $f_{i_j}(x)\in U_{i_j}$ for each $j\leq n$, from which it follows that there exists $N_j\in\mathbb{N}$ such that $f_{i_j}(x_n)\in U_{i_j}$ for all $n\geq N_j$. It is easy to see that for $N=\max_{j\leq n}N_j$ one has $x_n\in U$ for all $n\geq N$.
Now, in your context, $x_n\rightharpoonup x$ is equivalent to $\varphi(x_n)\to \varphi(x)$ for all $\varphi \in E^*$, while $Tx_n\rightharpoonup Tx$ is equivalent to $\psi(Tx_n)\to\psi(Tx)$ for all $\psi\in F^*$. Since you have the former by hypothesis, the latter follows: for $\psi\in F^*$, we have that $\psi\circ T\in E^*$, hence $\psi\circ T(x_n)\to \psi\circ T(x)$.
In order to prove that if $T\colon (E,\|\cdot\|)\to (F,\|\cdot\|)$ is continuous then $T\colon (E,\sigma(E,E^*))\to (F,\sigma(F,F^*))$ is continuous, you can use the next fact, stated in the context of the beginning of my answer:
Fact 2. For any topological space $Z$, a function $f\colon Z\to (X,\sigma(X,\mathcal{F}))$ is continuous if and only if $f_i\circ f\colon Z\to (X_i,\tau_i)$ is continuous for all $i\in I$.
The proof is routine, but you can check it below:
If $f$ is continuous, then $f_i\circ f$ is trivially continuous. Conversely, if $f_i\circ f$ is continuous for all $i\in I$, we take an open set $V\subset X$, say $V=\bigcap_{j\leq n}f_{i_j}^{-1}(V_{i_j})$ for $i_j\in I$ and $V_{i_j}\in \tau_{i_j}$, and notice that $f^{-1}(V)=f^{-1}\left(\bigcap_{j\leq n}f^{-1}_{i_j}(V_{i_j})\right)=\bigcap_{j\leq n}f^{-1}\left(f^{-1}_{i_j}(V_{i_j})\right)=\bigcap_{j\leq n}(f_{i_j}\circ f)^{-1}(V_{i_j})$.
Now, if $T\colon (E,\|\cdot\|)\to (F,\|\cdot\|)$ is continuous, then for any $\psi\in F^*$ one has $\psi\circ T\colon (E,\|\cdot\|)\to k$ continuous, but this implies that $\psi\circ T\colon (E,\sigma(E,E^*))\to k$ is continuous (by the definition of $\sigma(E,E^*)$). So, by replacing $Z$ by $(E,\sigma(E,E^*))$ in the Fact 2, we got the result.
Although it is not part of your question, I am almost sure that the converse is true, but I don't remember how to prove it.
You can easily prove this by using nets. Let $(x_i)$ be a convergent net and $x=\lim_i x_i$. Note that $x_i \rightarrow x$ iff $x'(x_i)\rightarrow x'(x)$ for all $x'\in X'$. Now, since $T$ is continuous, we have for all $y'\in Y'$ that $y'\circ T \in X'$.
It then follows for all $y'\in Y'$ that $y'(Tx_i)=(y'\circ T)(x_i)\rightarrow (y'\circ T)(x)=y'(Tx)$. Since $y'$ was arbitrary, $Tx_i \rightarrow Tx$ and $T$ is weak-weak-continuous.
Edit: This can also be done without nets. By definition of the weak topology, it suffices to show that $y'\circ T$ is continuous for all $y' \in Y'$. But this is clear, since $y'\circ T \in X'$.
Edit 2: Remember that the weak topology on $Y$ has the following universal property: For any topological space $T$ and any map $f:T \rightarrow Y$ it holds that $f$ is continuous iff $y'\circ f$ for all $y' \in Y'$. To see this, suppose $y' \circ f$ is continuous for all $y\in Y'$. Let $V\subset Y$ be open and $x\in f^{-1}(V)$. We will show $x$ is an inner point of $f^{-1}(V)$.
Now, since $V$ is open, by the definition of the weak topology there are functionals $y'_1,\ldots,y_n'$ and open sets $U_1,\ldots,U_n\subset \mathbb{K}$ such that $f(x)\in \left(\bigcap_{k=1}^n y'^{-1}_k (U_k)\right) \subset V$. It follows that $x\in \bigcap_{k=1}^n(y'_k\circ f)^{-1}(U_k)=f^{-1} \left(\bigcap_{k=1}^n y'^{-1}_k (U_k)\right) \subset f^{-1}(V)$. By assumption, $(y'_k\circ f)^{-1}(U_k)$ is open for all $k$, so we are done.
The linearity is not needed. In fact, what you have is a continuous function $T$ between topological spaces $X$ and $Y$ (they're normed but that's not relevant) and a convergent net (or sequence) $x_h \rightarrow x$ in $X$. Then show that $T(x_h)$ converges to $T(x)$. (weak topology not needed here.)
And this is quite easy: take an open set $O \subset Y$ that contains $T(x)$. Then by continuity of $T$, $T^{-1}[O]$ is open in $X$, and contains $x$. As $x_h \rightarrow x$, there is some $h'$ such that for all $h \ge h'$, we have that $x_h \in T^{-1}[O]$. But this means that for all $h \ge h'$ we have that $T(x_h) \in O$. So $T(x_h)$ converges to $T(x)$ by definition.
