As pointed out by Thomas, it follows from the definition of infimum.
A subset $A$ of the real numbers is said to be bounded below if there is $b \in \mathbb{R}$ such that $b \leq a$ for all $a \in A$. For such a set, the infimum of $A$ is the greatest lower bound for $A$ which is guaranteed to exist due to the completeness of $\mathbb{R}$.
As $\inf_{z\in K}\|x - z\| = d$ we know that $d \leq \|x - z\|$ for every $z \in K$. Now for each $n \in \mathbb{N}\setminus\{0\}$ let $y_n \in K$ be such that $d \leq \|x - y_n\| \leq d + \frac{1}{n}$. For each $n$, such a $y_n$ exists; if no such $y_n$ existed for a particular $n$, then we would have $d + \frac{1}{n} \leq \|x - z\|$ for all $z \in K$ which contradicts the fact that $\inf_{z\in K}\|x-z\|=d$.
So we have a sequence $(y_n)$ such that $d \leq \|x-y_n\| \leq d + \frac{1}{n}$. Taking limits we obtain $d \leq \displaystyle\lim_{n\to\infty}\|x-y_n\|\leq d$ and hence $\displaystyle\lim_{n\to\infty}\|x-y_n\| = d$.