[SOLVED] Closet Point/Sequentially compact subspace

Problem:

Let X be a metric space and let $\displaystyle K \subseteq X$ be a sequentially compact subspace. Let *p* be a point in *X*\*K*. Prove that there is a point $\displaystyle q_{0} $ that is closet to *p* of all points in *K* in the sense that

$\displaystyle d(p,q_{0}) \leq d(p,q)$ for all points *q* in *K*.

Is this point unique?

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Suppose that $\displaystyle K \subseteq X$ is sequentially compact subspace of the metric space *X*. Then *K* is a closed subset of *X*.

Define the function $\displaystyle f:K \rightarrow R$ by $\displaystyle f(q)=d(q,p)$ for all *q* in *K* and *p* be a point in *X*\*K*. I want to show that this function *f* is continuous first, but I don't know how nor what theorem to use.

Thank you for your time. Much appreciated.