Strictly increasing function from $\mathbb{R}$ into $\mathbb{R}\backslash\mathbb{Q}$ It seems this question is duplicated...Still I would be grateful for any hint to my second question.
Does there exist a strictly increasing function $f:\mathbb{R}\rightarrow\mathbb{R}\backslash\mathbb{Q}$, i.e., from real numbers to irrational numbers? It cannot be surjective, since otherwise one can show that it is continuous and obtain a contradiction, but without assuming surjectivity I have not found a contradiction. 
I do have a candidate function with that property by translating decimal expansion to continued fraction. More precisely, define a function $f:(0,1)\rightarrow\mathbb{R}\backslash\mathbb{Q}$ as follows. Let $x\in(0,1)$ with decimal expansion $0.a_1a_2a_3...$, define $f(x)$ to be $[0;1,1+a_1,1,1+a_2,...]$. As far as I understand infinite continued fraction must represent irrational number, and it seems $f$ preserves order. Is this a correct example?
More generally, what subset of reals can be the image of strictly increasing function?
 A: Your example is fine so long as you specify some way to disambiguate which decimal expansion to use for numbers with a terminating one. You could also solve this by taking the decimal digits and interleaving them with those of some irrational number - this would never yield a repeating decimal and would more obviously be increasing.
You can work out an answer to your more general question too:

A set $S$ is the image of an increasing function $f:\mathbb R\rightarrow\mathbb R$ if $S$ has at least two elements and for every point $x$ not in $S$, exactly one of the following holds:
  
  
*
  
*There is a maximum element in $S\cap (-\infty, x)$.
  
*There is a minimum element in $S\cap (x,\infty)$.
  
*The set $S\cap (-\infty, x)$ is empty.
  
*The set $S\cap (x,\infty)$ is empty.
  

Otherwise stated: each component of the complement of $S$ must either be a bounded half-open interval or an unbounded-on-one-side open interval and $S$ must not just be a singleton.
These conditions are clearly necessary using the first set of conditions: if $x$ is not in the image of an increasing function, then $f^{-1}((-\infty,x))$ and $f^{-1}((x,\infty))$ are disjoint and have union $\mathbb R$ - so either one of these sets is empty, or one of them includes a boundary point, leading to one of the conditions listed.
These conditions are also sufficient, but this direction is harder: the clean way to do this is to axiomatize $\mathbb R$ as a total order and then show that $S$ satisfies the same axioms. An appropriate axiomatization would be as follows:


*

*$R$ has at least two elements.

*$R$ has a countable dense subset $Q$ in the sense that for any $x,y\in R$ with $x<y$ there exists some $z\in Q$ such that $x<z<y$.

*Every non-empty bounded subset of $R$ has a supremum.

*$R$ does not have a minimum or maximum element.
You can note that $Q$ has to be order-isomorphic to $\mathbb Q$ since it is a countable dense order without minimum or maximum, then see that each element of $R$ is uniquely the supremum of some non-empty closed bounded interval of $Q$ - which one recognizes as the completion of $Q$ by Dedekind cuts. This implies that $R$ is order isomorphic to $\mathbb R$.
Thus, if we check that $S$ satisfies these conditions, we are done. First, let $\mathscr C$ be the set of bounded components of $\mathbb R\setminus S$. Each of these is a half-open interval $I$, with exactly one endpoint $e_I$ in $S$. Note that $\mathscr C$ is at most countable because each element thereof has positive length and is disjoint from every other element. Let $E=\{e_I : I\in\mathscr C\}$ be the set of these elements and let $Q = E \cup (S\cap \mathbb Q)$.
Claim: $Q$ is order-dense in $S$.
Proof: Let $x,y\in S$ satisfy $x<y$. Choose any rational $z$ such that $x<z<y$. If $z\in S$ then $z\in Q$ so we are done. Otherwise, let $I\in\mathscr C$ be the interval of the complement of $S$ containing $z$. If $e_I$ is between $x$ and $y$ we are done - otherwise, $e_I$ must equal either $x$ or $y$. In the former case, we have $I=(x,x']$ with $x' < y$. Note too that $x'$ must be the infimum of $S \cap (x,\infty)$ - so we may choose some $s\in S$ such that $x' < s < y$ since $y$ must not be a lower bound of $S \cap (x,\infty)$. If $(x',s)\subseteq S$, we may choose any rational in that interval and observe that it is in $(x,y)\cap Q$. Otherwise, we can choose some $z'\in (x',s)\setminus S$ and let $I'$ be the interval of $\mathscr C$ containing it. Then $e_{I'}$ must be in $[x',s]$ so $x<e_{I'}<y$ in this case. The case where $e_I=y$ proceed similarly by reflecting the argument.
Claim: Every non-empty bounded subset of $S$ has a supremum.
Proof: Let $B\subseteq S$ be any such subset of $S$ and let $b\in B$ be some element of $B$ and let $u\in S$ be any upper bound to $B$. Let $x\in\mathbb R$ be the supremum of $B$ as a subset of $\mathbb R$. If $x\in S$, then we are done because $x$ must be the supremum of $B$. Otherwise, let $I\in\mathscr C$ be the interval of the complement containing $x$. Note that $x$ must be an endpoint of $I$, since any interval containing $x$ on the interior would contain some element of $B$ as well. In particular, the interval must be of the form $I=[x,x')$ for some $x'>x$. Then $x'\in S$ is the supremum to $B$ in $S$, since every upper bound to $B$ must be at least $x$ and the least such value in $S$ is $x'$. We have therefore found the supremum.
Claim: $S$ has no minimum or maximum.
Proof: Let $x\in S$. Choose any $y \in \mathbb R$ less than $x$. If $y\in S$, we are done. Otherwise, let $I$ be the component of $\mathbb R\setminus S$ containing $y$. If $I$ is bounded below, then there must be some $z\in S$ with $z<y<x$, so $x$ is not a minimum. Otherwise, $S\cap (-\infty, y)$ is empty, therefore, by the condition, there is no minimum element in $S\cap (y,\infty)$ - also meaning that $x$ is not a minimum. This logic can be repeated symmetrically to see that $x$ is also not a maximum. As $x$ was arbitrary, we are done.
Together, these claims show that $S$ satisfies the order-theoretic axioms that uniquely define $\mathbb R$, hence must be order-isomorphic to $\mathbb R$, meaning there is an increasing bijective map from $\mathbb R$ to $S$. Inherently, this proof is constructive, since the order isomorphism between countable dense orders without a minimum or maximum may be determined constructively and then this order isomorphism will be extended in some unique way to the desired one - but the details of doing this are incredibly messy and, usually, you don't need such complicated machinery for specific cases you might be interested in.
Note also that it is easy to construct some such $S$ avoiding any measure $0$ set you desire: choose an open set $U$ with finite measure covering the set you wish to avoid. Let $\bar U$ be $U$ union the lesser endpoint of each interval composing $U$. Then $S\setminus \bar U$ satisfies the conditions.
A: I think your example is good, just choose either of the decimal expansions when you have a choice.  Then $x< x'$ implies that the expansion of $x$ precedes the expansion of $x'$ in lexicographic ordering and you can apply your transformation to continued fractions. 
Here is a similar trick that only uses (binary) expansions. Consider for 
every $x \in (0,1]$ the infinite binary  expansion
$$x = \sum_{n\in D_x}\frac{1}{2^n}$$
Now take an increasing map $\phi\colon \{1,2,\ldots,\}\to \{1,2,\ldots\}$
and define
$$f(x) = \sum_{n\in D_x}\frac{1}{2^{\phi(n)}}$$
If $\phi(n) = n^2$, $f$ has all values irrational, while if say $\phi(n) = n!$ then $f$ has all values Liouville numbers. 
