Assume a set in $L^2$, $$A:= \{u ∈ L^2(\Omega) : a ≤ u(x) ≤ b \;\;a.e\}$$ where $\Omega \subset \mathbb{R}^n$ open and bounded. Is $A$ weakly sequentially closed? Is it bounded?
I have shown boundedness as follows:
To be bounded $A$ should have a finite diameter.
$$d(A)=max(||u_1-u_2||: u_1,u_2 \in A)\\ =max(||u_1||+||u_2||)= finite$$
as $u_1$ ,$u_2 \in L^2(\Omega)$.
Since, strong convergence $f_n\to f$ in this space implies existence of subseqence $f_{n_k}$ which is convergent to $f$ almost everywhere, the set $A$ is closed in strong topology. It is easy to observe that's $A$ is convex. But in any Banach space strongly closed and convex set is weakly closed. Since in any topological space a closed set is sequentially closed, therefore the set $A$ is weakly sequentially closed.
Your proof of boundedness seems circular. That for $u \in A$ the norm $\left\lVert u \right\rVert$ is finite, is what one needs to prove. We can show that
$$ \newcommand{\norm}[1]{\left\lVert#1\right\rVert} \norm{u}_{L^{2}(\Omega)} = \left( \int_{\Omega} |u(x)|^{2}dx \right)^{\frac{1}{2}} < \infty $$
by bounding
$$ \norm{u}_{L^{2}(\Omega)}^{2} = \int_{\Omega} |u(x)|^{2} dx \leq |\int_{\Omega} u(x)^{2} dx| \leq \int_{\Omega} \max{ \lbrace a^{2}, b^{2} \rbrace } dx = C \mu(\Omega) $$
That the set $$ A = \lbrace u \in L^{2}(\Omega) : a \leq u(x) \leq b \text{ a.e.} \rbrace $$ is closed can be shown, perhaps more clearly, by using a contradiction.
Let $(u_{n})_{n} \subset A$ and $u_{n} \rightarrow u$ as $n \rightarrow \infty$ in $L^{2}$ norm. And suppose that $u \not\in A$, meaning that there exists a subset $B \subset \Omega$ with $\mu(B) > 0$ where, without loss of generality, $u(x) > b$. Then, for some $\epsilon > 0$ and since $u_{n} \in A$ \begin{align*} \norm{u - u_{n}}_{L^{2}(\Omega)}^{2} &= \int_{\Omega} |u(x) - u_{n}(x)|^{2} dx \\ &= \int_{B} | \underbrace{u(x)}_{> b + \epsilon} - \underbrace{u_{n}(x)}_{\leq b} |^{2} dx + \underbrace{ \int_{\Omega \setminus B} |u(x) - u_{n}(x)|^{2} dx }_\text{$u_{n} \rightarrow u$ in $L^{2}$ pointwise} \\ &\geq \epsilon^{2} \mu(B) + 0 \\ &> 0 \end{align*} which is a contradiction since $u_{n} \rightarrow u$ in $L^{2}$. Therefore $u \in A$.
