The most usual way to show that a topology $A$ is stronger than a topology $B$ is to show that a convergent net in $A$ is also convergent in $B$. You show that the two topologies are equal by showing that both are stronger than the other.
Since all the topologies you want to consider are linear, it is enough to consider convergence at $0$.
Here, suppose that $T_j\to 0$ in norm, i.e. $\|T_j\|\to0$. For any $x\in H$,
$$
\|Tx\|+\|T^*x\|\leq\|T\|\,\|x\|+\|T^*\|\,\|x\|=2\|T\|\,\|x\|\to0;
$$
so norm convergence implies strong* convergence.
From $\|Tx\|\leq\|Tx\|+\|T^*x\|$ we get that strong* convergence implies strong convergence.
If $T_j\to0$ strongly, this means that $T_jx\to0$ for all $x\in H$. Then
$$
|\langle T_jx,y\rangle|\leq\|T_jx\|\,\|y\|\to0,
$$
so strong convergence implies weak convergence.
So far we have shown the "weaker than" implications. Now we need to see that they are strict.
Fix an orthonormal basis $\{e_j\}$ of $H$ and let $E_j$ be the projection onto the span of $e_j$ (i.e. $E_jx=\langle x,e_j\rangle e_j$). Then for any $x=\sum_j\alpha_je_j$,
$$
\|E_jx\|+\|E_j^*x\|=2\|E_jx\|=|\alpha_j|\to0,
$$
so $E_j\to0$ in the strong* topology. But $\|E_j\|=1$ for all $j$, so $\{E_j\}$ does not converge in norm.
Now consider the left-shift operator $T$ given by $Te_1=0$, $Te_j=e_{j-1}$ for $j>1$ (we don't need to assume $H$ separable here, just use a well-ordering of the index set). If you are familiar with it, $T$ is the adjoint of the unilateral shift. Put $T_n=T^n$, $n\in\mathbb N$. Then
$$
\|T_nx\|^2=\sum_{k>n}|\alpha_k|^2\to0\ \ \text{ as }n\to\infty;
$$
so $T_n\to0$ in the strong topology. But $T^*$ is an isometry, so
$$
\|T_nx\|+\|T_n^*x\|\geq\|T_n^*x\|=\|x\|
$$
for all $n$, so $\{T_n\}$ does not converge in the strong* topology.
Finally, we need a net that converges weakly but not strongly. Here we can use the unilaterial shift $S$ ($T^*$ above). If $x=\sum\alpha_ke_k$, $y=\sum_j\beta_je_j$, then
$$
|\langle S^nx,y\rangle|=|\sum_k\alpha_k\beta_{k+n}|\leq\sum_k|\alpha_k|\,|\beta_{k+n}|\leq\left(\sum_k|\alpha_k|^2\right)^{1/2}\,\left(\sum_k|\beta_{k+n}|^2\right)^{1/2}\to0
$$
(the second sum goes to zero because of the $n$). So $S^n\to0$ weakly but not strongly (recall that $S$ is an isometry).
I think the answer is no. At the very least, it is not true that the set of closed sets is the complement of the set of open sets. Instead, the closed sets are the complements of the open sets, which is different. Indeed, adding open sets also adds closed sets!
The example I have in mind is $X=L^2([0,\pi])$, where $\sin(nx)$ is a sequence which is contained on a sphere and converges weakly to zero, which is not on the sphere. Thus I think that the sphere cannot be weakly closed, but it is certainly strongly closed (the norm is strongly continuous and singletons are closed in $\mathbb{R}$).
Best Answer
To show that the weak topology is strictly weaker, you need a norm-open set that is not weakly open. The unit ball $B=\{x:\|x\|<1\}$ is such a set. Indeed, suppose it is weakly open. By the definition of the weak topology, there exists a finite set of functionals $f_1,\dots,f_n$ such that $$ \{x: |f_k(x)|<1 \text{ for }k=1,\dots,n\} \subset B $$ But from any $n+1$ linearly independent elements $x_1,\dots,x_{n+1} \in X$ you can make a nonzero linear combination $x_0$ such that $f_k(x_0)=0$ for all $k=1,\dots,n$. (Think of the rank of the matrix $(f_i(x_j))$.) It follows that $tx_0\in B$ for all $t\in\mathbb{R}$, which is impossible.