Let $H$ be a complex Hilbert space and $\mathcal B(H)$ be the set of bounded linear operators on $H$. Let $E$ be a nontrivial projection and consider $\mathcal{A}=\{\alpha E:\alpha\in \mathbb C\}$. Prove: $\mathcal{A}$ is weakly closed in $\mathcal B(H)$.
Recall that the weak topology $\tau_w$ on $\mathcal B(H)$ is defined by semi-norms $\mathcal P=\{p_{xy}: x,y\in H\}$, where $$p_{xy}(B)=|(Bx,y)|, \qquad \forall B\in\mathcal B(H).$$
I want to show that $\mathcal B(H)\setminus \mathcal{A}$ is open in $(\mathcal B(H),\tau_w)$. Let $B\in \mathcal B(H)\setminus \mathcal{A}$, so we need to find $\epsilon>0$ and finite elements $\{x_i,y_j:1\leq i\leq m,1\leq j\leq n\}\subset H$ such that $$B+\bigcap_{1\leq i\leq m,1\leq j\leq n}\{C\in \mathcal{B}(H): |(Cx_i,y_j)|<\epsilon\}\subset \mathcal B(H)\setminus \mathcal{A}. $$
This looks messy and I don't know how to continue now. Any help will be appreciated.
A comment first: what you are using is the weak operator topology. The weak topology is something else. While is true that it is common to say that something converges "weakly" to mean "in the weak star topology", no one would use the name "weak topology".
From a more abstract point of view, you have a finite-dimensional subspace, and in a finite dimensional subspace there is a single locally convex topology, and the subspace is closed in it.
If you want a direct proof, the easiest way is using convergence. Since $B(H)$ is complete in the weak operator topology, all you need to show is that if $\{\alpha_jE\}_j\to T$ then $T=\alpha E$ for some $E$. For any $x\in H$, you have $$ \langle Tx,x\rangle=\lim_j\langle \alpha_jEx,x\rangle=\langle Ex,x\rangle\,\lim_j\alpha_j. $$ Using $x$ such that $Ex\ne0$, we get that the net of numbers $\{\alpha_j\}$ converges to some $\alpha\in\mathbb C$, showing that $\langle Tx,x\rangle=\langle \alpha Ex,x\rangle$ for all $x\in H$. This shows that $T=\alpha E$ and so your set is closed.