Let $X$ be a normed space. Assume $Y\subset X$ is a linear subspace of $X$. Prove that $$\overline{Y}=\bigcap_{f\in X^*,\mathrm{ker}f\supset Y}\mathrm{ker}f$$
I was able to prove that $\overline{Y}\subseteq\bigcap_{f\in X^*,\mathrm{ker}f\supset Y}\mathrm{ker}f$. However, I couldn't find a way to show the converse.
My attempt: Obviously, a direct way to attack the problem it would be consider a $y\in \cap_{f\in X^*,\mathrm{ker}f\supset Y}\mathrm{ker}f$ and construct a sequence of $\{y_n\}_n\subset Y$ such that $y_n\rightarrow y$ in norm. I was thinking in the following identity $$\lvert\lvert y_n-y\rvert\rvert=\sup_{f\in X^*,\lvert\lvert f\rvert\rvert\leq1}|f(y_n-y)|.$$ Since the qualities of $y$ are stated in terms of bounded functionals, but my attempt was unfruitful.
Is there any suggestion or hint to approach the exercise?
I'll turn my comment into an answer.
Using the second importance consequence on Wikipedia: we see that for each $w \in V \setminus \overline{Y}$. There exists a linear functional that vanishes on $Y$ and $f(w)=1 \notin \ker f$, which implies that $w \notin \bigcap \ker f$. Hence, it follows that their intersection is contained in the closure of $Y$.
On SE, the second consequence is proven here.