Hilbert space projection theorem

To prove: If for every $$f\in H$$ ($$H$$ is a Hilbert space) there is a $$p\in V$$ such that $$\|p−f\|=\min_{v\in V}\|v−f\|$$ then $$V$$ is closed.

I was able to prove the converse of this statement, but not this. I am not able to write down the proof coherently, and A subspace $X$ is closed iff $X =( X^\perp)^\perp$ is confusing me even more! Can someone write down steps of hints to complete this proof?

Well, since a Hilbert space is in particular a metric space, $$V$$ being closed can be characterized by convergent sequences.
That is, suppose $$v_n \in V$$ converges to some $$f \in H.$$ Then we have that there is some $$p \in V$$ such that $$\|p-f\| = \min_{v\in V} \|v-f\| \leq \lim_{n\to\infty}\|v_n-f\| = 0 \\ \implies f = p \in V$$