The question is whether rationality is implied by continuity and monotonicity. To show that this is not the case, a counterexample would suffice. We are therefore looking for an intransitive, incomplete, monotone, continuous preference relation.
Suppose $X=\{x\geq 0,y\geq 0:x+y=1\}$. Thus, we form preferences over points of a line from $(0,1)$ to $(1,0)$.
Consider the preference relation defined by $(1,0) \succ (.5,.5) \succ (0,1) \succ (1,0)$ which is incomplete otherwise.
Rationality
Rationality consist of completeness and transitivity of the preference relation, defined as follows:
Completeness
A preference relation is complete, if for all $x,y \in X$, we have $x\succsim y$, $y\succsim x$, or both.
$(.5,.5)\not \succsim (.5,.5)$, thus the preference relation is not complete.
Transitivity
A preference relation is transitive, if $x\succsim y$ and $y \succsim z$ imply $x\succsim z$.
$(1,0)\succsim (.5,.5)$ and $(.5,.5) \succsim (0,1)$ hold but $(1,0)\not \succsim (0,1)$, thus the preference relation is not transitive.
Continuity
A preference relation is continuous if for all sequences ${(x_i,y_i)}_{i=1}^{\infty}$ converging to $(x,y)$ with $\forall i: x_i \succsim y_i$ we have $x \succsim y$.
The preference relation does not violate continuity. Consider a sequence $x_i \succsim y_i$ which converges to $x,y$. These sequences can only be such that $x_i=x$ and $y_i=y$, and $x\neq y$, since all other $x_i,y_i$ either do not converge to $x,y$, or do not fulfill $x_i \succsim y_i$. But clearly if $x_i\succsim y_i$ then $x\succsim y$.
Monotonicity
A preference relation is monotone, if $x \geq y$ implies $x \succsim y$.
The $\geq$ relation considers all elements of $X$ incomparable, thus the preference relation is monotone.
Thus, we have an intransitive, incomplete, monotone, continuous preference relation.