Is $\big\{x \in L^2[0,1]:\int_0^1x(t)\,dt=0\big\}$ a closed set?

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I have been stuck with this problem for two days, can I have hint to prove whether \begin{equation} Y=\big\{ x \in L^2[0,1]:\int_0^1 x(t)\,dt=0 \big\} \end{equation} is a closed subset of $L^2[01,]$ or not?

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YES.

This is due to the fact that the functional $\varphi : L^2[0,1]\to\mathbb R$, with $$ \varphi(\,f)=\int_0^1f\,dx $$ is bounded (and hence continuous), since $$ |\varphi(\,f)|\le\int_0^1|\,f|\,dx\le \left(\int_0^1|\,f|^2\,dx\right)^{1/2}, $$ and therefore $\varphi^{-1}[\{0\}]$ is a closed set.