There is a given function $\mathbf{a}:[0,1]^2 \rightarrow [0,1]^2$. For any given $n\in \mathbb{N}$, define $x_k =y_k= \frac{k}{n}, k \in \{0,\cdots,n\}$.
Define:
\begin{align*} I_n&=\{\mathbf{a} \in ([0,1]^2)^{[0,1]^2}: t,t' \in [x_{k_x},x_{k_x+1}] \times [y_{k_y},y_{k_y+1}]\\ &\text{for some}\;k_x,k_y \in \{0,\cdots,n\}\\ &\implies \mathbf{a}(t) = \mathbf{a}(t') \} \end{align*}
$I_n$ is clearly non-empty for any $n \in \mathbb{N}$.
I have to construct a sequence $\{\mathbf{a}_n\}^\infty_{n=1}$, $\mathbf{a}_n \in I_n\;\forall\;n$, which converges to $\mathbf{a}$ in the $L^2$-norm, where the $L^2$-norm is defined by ($\lambda$ is the Lebesgue measure on $[0,1]^2$):
$$ \|\mathbf{a}\|:=\left(\int\limits_{t \in [0,1]^2} \left(\left|a_1(t)\right|^2+\left|a_2(t)\right|^2\right)d \lambda\right)^{1 / 2} $$
My attempt:
For any given $\mathbf{a}$ and $n \in \mathbb{N}$, pick:
\begin{equation*} \mathbf{a}_n \in \arg\min\limits_{\mathbf{a}' \in I_n} \lVert \mathbf{a}-\mathbf{a}' \rVert \end{equation*}
I have shown (hoping it's correct!) that $I_n$ is (isomorphic to) a closed subset of a Euclidean space -- namely, $([0,1]^2)^{n^2}$ -- so the above is well-defined.
Clearly, $\lVert \mathbf{a}-\mathbf{a}_n \rVert $ is decreasing in $n$. But I'm not able to clearly construct the argument that for any $\epsilon>0$ there exists an $n$ such that for all $n'>n$, $\lVert \mathbf{a}-\mathbf{a}_{n'} \rVert <\epsilon$. Any help is most appreciated.