Prove that there exists $y_0\in C$ such that $d(x,y)=\inf_{y\in C} d(x,y)$, i.e. $y_0$ is a closest point to $x$ in $C$. 
If $C$ is a closed subset of $R^n$ and $x\in R^n$, prove that there exists $y_0\in C$ such that $d(x,y)=\inf_{y\in C} d(x,y)$, i.e. $y_0$ is a closest point to $x$ in $C$.

Here's what I got but I'm not sure how to finish the last step:
First we notice that the set $\{d(x,y):y\in C\}$ is in $R$ and it is bounded below, so the infimum of that set actually makes sense. Denote $\inf_{y\in C} d(x,y)$ as $L$.
Since $L$ is the infimum, then $L+\frac{1}{n}$ is not the infimum. Thus for each $n$, there is an $d_n$ such that $L<d_n<L+\frac{1}{n}$. We know that $R$ is complete, so $C$ is complete as well since it is closed. So I think I really want a Cauchy sequence in $C$. Then I would argue that it converges to a point in $C$ and then show that at that point we achieve the infimum.
So I tried consider $B(x;d_n)\cap C$.For each $n=1,2,3...$, we take $y_1,y_2,y_3,...$ in that intersection. We would never have empty set because $d_n> L$ (now sure how to formally write this part?). Also I think this sequence is Cauchy because given any $\frac{1}{n}$, there will be infinitely many points between $L$ and $L+\frac{1}{n}$. (not sure...)
 A: Pick some $y_1 \in C$, and define $r=d(x,y_1)$. Let $B$ denote the closed ball of radius $r$ centered at $x$. Then $C \cap B$ is nonempty (because it contains $y$), and more importantly $C \cap B$ is compact: it is closed because it is the intersection of two closed sets, and it is bounded because it is contained in the bounded set $B$.
Define $f: C \cap B \to \mathbb{R}$ as $y \mapsto d(x,y)$. Then $f$ is continuous, and so it achieves its minimum value on the compact set $C \cap B$. Therefore there exists some $y_0 \in C \cap B$ such that $f(y_0) = \inf_{y \in C \cap B} f(y)$. Easily enough, $y_0$ is the point in $C$ which is closest to $x$.
A: I am assuming that $d$ is a metric induced by a norm, that is, $d(x,y) = \|x-y\|$.
Let $y_n \in C$ be such that $\lim_n d(x,y_n) = \inf_{y \in C} d(x,y)$. We see that $y_n$ is bounded, hence there is a subsequence $y_{n_k}$ and a point $y$ such that $y_{n_k} \to y$. Since $C$ is closed, we have $y \in C$, and
since $d$ is continuous, we have $\lim_n d(x,y_n) = d(x,y)$.
Addendum:
Let $\alpha = \inf_{y \in C} d(x,y)$. By definition of $\inf$, for all $\epsilon>0$ there is some $y \in C$ such that $\alpha \le d(x,y) < \alpha+\epsilon$. Choosing $\epsilon= {1 \over n}$ gives the sequence $y_n$.
To see why $y \mapsto d(x,y)$ is continuous, note that
$d(x,y) \le d(x,y')+d(y',y)$ which gives $d(x,y) -d(x,y') \le d(y',y)$,
reversing the roles of $y,y'$ shows that $|d(x,y) -d(x,y')| \le d(y',y)$,
from which continuity follows.
