The continuity axiom can be interpreted as follows: Let $(\textbf{x}^n)$ and $(\textbf{y}^n)$ be two sequence of goods, such that $(\textbf{x}^n)$ converges to $\textbf{x}$ and $(\textbf{y}^n)$ to $\textbf{x}$. Then the continuity axiom says that if $\forall n$ $(\textbf{x}^n)$$\succsim$$(\textbf{y}^n)$, then $\textbf{x}$$\succsim$$\textbf{y}$ also holds true(here, $(\textbf{x}^n)$,$(\textbf{y}^n),\textbf{x}, \textbf{y} \in\mathbb{R_+^n} $). This statement is equivalent to the fact that the upper contour set and lower contour set is closed (think of $(\textbf{x}^n)$ and $(\textbf{y}^n)$ to be convergent sequences of the upper and lower contour sets respectively,i.e;$(\textbf{x}^n)\in (upper contour)$, and $(\textbf{y}^n)\in (lower contour)$, $\forall n$. Then, the sequences should converge within their respective sets).
The above statement, however, does not hold true for lexicographic preferences. To see that, consider any two bundles of $\mathbb{R_+^2} $represented by $((1/2)^n,0) \& (0,k)$ where, $k>0$. It is routine to check that the bundle $((1/2)^n,0)$ converges to $(0,0)$. Now, according to the concept of lexicographic preference(I hope you're familiar with it), $((1/2)^n,0)\succsim(0,k)$. However, the preferences change as the bundle $((1/2)^n,0)$ converges to $(0,0)$(i.e; now the bundle $(0,k)\succsim(0,0)$,as n tends to an infinite value). Thus, lexicographic preferences violate the continuity axiom.
