A monotonic function that intersect with all lines in $\mathbb R^2$ 
Let $f:\mathbb R\to\mathbb R$ be a monotone function. Let $\gamma=\{(x,y)\ |\ y=f(x)\}$ is a curve in $\mathbb R^2$.
Does there exists a $f$ such that $\gamma\cap L\neq \emptyset \ \forall L\subset\mathbb R^2$; that is, $\gamma$ intersects with all lines?


Intuitively there must exist some $f$; imaging a "ladder" bouncing between $y=x$ and $y=1.1x$. But I failed to rigoroze this.
What is the sufficient and necessary condition for a curve that does not intersect with at least one line (which means: $\exists L\subset\mathbb R^2$ such that $\gamma\cap L=\emptyset$)?
 A: 
Let $f:\Bbb R\to\Bbb R$ be a continuous function.
  If $f (x)/x\to +\infty$ as $x\to\pm\infty $, then the graph $\gamma $ of $f $ meets all lines of the plane.

Clearly $\gamma $ meets all vertical ones.
If $f (x)/x\to+\infty$ as $x\to\pm\infty $ then
$$f (x)-mx-q\sim f (x)=
\begin{cases}
+\infty&x\to+\infty\\
-\infty&x\to-\infty
\end{cases}$$
as $x\to\pm\infty $, hence $f (x)-mx-q$ has at least a root by intermediate values theorem.
Example: $f (x)=x^3$

Let $f:\Bbb R\to\Bbb R$ be a continuous function.
  If $f $ is (upper and lower) unbounded and $f (x)/x\to 0$ as $x\to\pm\infty $, then the graph $\gamma $ of $f $ meets all lines of the plane.

Since $f $ is unbounded, its graph meets all orizontal lines.
On the other hand, for $m\neq 0$
$$f (x)-mx-q\sim -mx$$
as $x\to\pm\infty $, hence $f (x)-mx-q$ as at least a root by intermediate value theorem.
Example: $f (x)=\sqrt [3]x $.

Let $f:\Bbb R\to\Bbb R$ be continuous.
  If there exists a line which doesn't meets the graph $\gamma$ of $f$, then there exists infinitely many lines which doesn't meet the graph $\gamma$ of $f$.

For if $f(x)-mx-q\neq 0$ for every $x\in\Bbb R$, then $f(x)-mx-q>0$ or $f(x)-mx-q<0$ for every $x\in\Bbb R$.
If wlog, $f(x)-mx-q>0$ for every $x\in\Bbb R$, then $f(x)-mx-q'>0$ for every $x\in\Bbb R$ and $q'<q$.
A: If we restrict the problem for only "nice" monotonic functions $f$, which have the derivative in $\mathbb R$, then such $f$ must fulfill two conditions:


*

*It has neither upper, no lower bound.





*Its derivative $f'$ has no upper bound if $f$ is increasing.
(No lower bound for $f$ decreasing.)



The pictures show simple counter-examples with disjunct line $a$, when the function $f$ don't fulfill the corresponding condition.
In other words, we have to get rid of disjunct lines $y = kx+q$ - both for $k=0$ and $k \ne 0$.
(Al lines $x=c$ intersect your curve - the graph of $f$.)

What is the sufficient and necessary condition for a curve that does not intersect with at least one line (which means: $\exists L\subset\mathbb R^2$ such that $\gamma\cap L=\emptyset$)?

For "nice" functions which have derivative in $\mathbb R$, the opposite of conditions "1. and 2.", i.e. "not 1 or not 2".
A: Concerning your first question: What you are looking for is a monotone function $f\colon\Bbb R \to \Bbb R$ such that for any $a,b\in\Bbb R$ the equation
$$ f(x)-(ax+b)=0$$
has at least one solution $x\in\Bbb R$. As Fabio Lucchini has already pointed out in the comments, $f(x)=x^3$ will do the trick, just like any other monotone polynomial function with uneven degree greater than 1.
