$\textbf{Problem Statement :}$ Let $K$ be a compact subset of $\mathbb{R}^n$, and suppose that $x \in \mathbb{R}^n$ does not lie in $K$. Let $d := inf(d(x,y) \space \vert \space y \in K)$ where $d(x,y)$ is the distance from $x$ to $y$. Show that $\exists \space z \in K $ so $d(x,z) = d$.
Consider the set $D = (d(x,y) \space \vert \space y \in K)$. d is a limit point of this set. This is noted by the definition of infimum, or d is the largest lower bound. This means that we are able to find a y such that $d(x,y) < d + \epsilon$, or $\vert d(x,y) - d \vert < \epsilon$. We may then choose a sequence of $y_n$ which converge to a $z$ such that $d(x,z) = d$, and since $K$ is compact, $z \in K$. Is this close to a proof? I didn't even use that $K$ is bounded, only that it is closed and I feel its incomplete.
Choose an approximate sequence $(y_{n})\subseteq K$ such that $d(x,y_{n})\rightarrow d$. As $K$ is sequentially compact, then there is a convergence subsequence $(y_{n_{k}})$ such that $y_{n_{k}}\rightarrow z\in K$, so $d(x,y_{n_{k}})\rightarrow d(x,z)$ and hence $d(x,z)=d$.
Your strategy is closed to choosing an approximate sequence.