Is the subspace of $L^2([0,1])$ of all "functions" vanishing on $[0,1/2]$ closed? I am trying to understand the following example from my lecture notes:

$\mathcal{H} := L^2([0,1],\lambda)$,
  then 
  $$K := \{f \in \mathcal{H} \colon f(x) = 0, \text{ for } 0 \leq x \leq \frac{1}{2} \;\lambda\text{-almost everywhere}\}$$
  is a closed subspace.

The subspace properties follow from Minkowski's inequality, i.e. for $[f],[g] \in K$
$$ \left(\int_{[0,\frac{1}{2}]} |f + g|^2 \right)^{1/2}
= \| (f+g)\cdot \chi_{[0,\frac{1}{2}]} \|_2
\leq \| f \cdot \chi_{[0,\frac{1}{2}]} \|_2 + \| g \cdot \chi_{[0,\frac{1}{2}]} \|_2
= 0
$$
Is this correct?
I'm not able to show closedness: Consider a sequence $([f_n])_{n \in \mathbb{N}}$ in $K$ converging to some $[f] \in \mathcal{H}$. The problem is that convergence w.r.t. the Norm in L2 only gives my a subsequence $(f_{n_k})$ (note that I dropped the equivalence classes) converging pointwise almost everywhere. So I'm guessing the proof can't be done using the dominated convergence theorem.
How would you prove closedness?

Yes. Assume not. Then $f_n\to f$ in $L^2$ but $f\neq 0$ on a set $E\subset [0,1/2]$ of positive measure. Then $\|f-f_n\|_{L^2(E)}=\|f\|_{L^2(E)}= C, \forall n$, which contradicts the fact that $f_n\to f$.
Edit: to be precise, the contradiction follows from what I said together with the fact that $\|\cdot\|_{L^2([0,1/2])}\geq\|\cdot\|_{L^2(E)}$.
