The Context
The origin of my question is my own answer to this question, where the continuity of the function $f: [0,1) \to [0,1)$ that only preserves the odd digits of its input value is analyzed.
In this context, I will slightly modify the definition given there, as follows.
The function I am considering now is $f: [0,+\infty) \to [0,+\infty)$ that preserves only the even digits of the input. So if $x$ has decimal expansion
$$x = \sum_{k=-\infty}^{+\infty}a_k 10^k,$$
then
$$f(x)=\sum_{k=-\infty}^{+\infty}a_{2k}10^{k}.$$
So for example
$$f(5\textbf{8}7\textbf{4}1\textbf{2}.7\textbf{8}0\textbf{3}4\textbf{0}5\textbf{1})=842.8301.$$
With a very similar approach to the one given here, it can be shown that $f$ is continuous almost everywhere (the only exception being the numbers whose least significant digit occupies an odd position), and right-continuous everywhere.
Edit. I am assuming, as in the original question, to adopt, in case of ambiguity, the finite version of the number's decimal expansion.
Note the self-similarity
$$f\left(10^{2k} x\right)=10^k f(x), \ \ \forall k\in \Bbb Z.$$
Below an approximate plot of the function $f$ in the range $[0,1]\times[0,1]$.
Red dots represent points actually belonging to the graph of $f$. The yellow line represents the graph of
$$s(x) = \sqrt x.$$
By self-similarity, a scaling of $10^{2k}$ of the $x$-axis and of $10^{-k}$ of the $y$-axis, for any $k\in \Bbb Z$, would give an exact replica of the given plot.
Introductory Observation
Aside from the trivial intersections between $f$ and $s$, that is all the points with coordinates
$$\left(10^{2k},10^k\right),$$
there are many other interesting intersections, such as (limiting ourselves to the range shown in the picture)
$$(0.25,0.5),$$
$$(0.36,0.6),$$
$$(0.0121,0.11),$$
and 'trikiest' ones, such as
$$(0.5776,0.76),$$
or even
$$(0.35295481,0.5941).$$
The Question
Is there any non-terminating decimal (or even irrational) $x$ such that
$$y=f(x) = s(x),$$
that is, is there any non-terminating decimal $y$ whose square contains – in the even positioned digits – the digits of the original number $y$?
Edit. I emphasize again that no infinite sequences of $9$'s are allowed, since we are adopting the finite decimal expansion version of the number, if this ambuiguity arises.
Best Answer
The answer is on the affirmative.
Consider the sequence \begin{eqnarray} \alpha_0 &=& 1,\\ \alpha_1 &=& 10005,\\ \alpha_2 &=& 1000505,\\ \alpha_3 &=& 10005050005,\\ \alpha_4 &=& 1000505000500000005,\\ \dots & &, \end{eqnarray} where, for $n>1$, $\alpha_{n+1}$ is obtained from $\alpha_n$ by appending a sequence of $2^{n-1}-1$ zeros and then a $5$: $$\alpha_{n+1} = \alpha_n \cdot 10^{2^{n-1}}+5.$$
Let us first show, by induction, that $$f\left(\alpha_n^2\right) = \alpha_n.\tag{1}\label{eq1}$$ Suppose that \eqref{eq1} holds true for a given $n$. Then \begin{eqnarray} f\left(\alpha_{n+1}^2\right) &=& f\left(\left(\alpha_n \cdot 10^{2^{n-1}}+5\right)^2\right)=\\ &=& f\left(\alpha_n^2\cdot 10^{2^n}+\alpha_n\cdot 10^{2^{n-1}+1}+25\right). \end{eqnarray} Note that
As a consequence
\begin{eqnarray} f\left(\alpha_{n+1}^2\right) &=& f\left(\alpha_n^2\cdot 10^{2^n} + 25\right) =\\ &=&\alpha_n \cdot 10^{2^{n-1}} + 5=\\ &=& \alpha_{n+1}. \end{eqnarray}
Consider now the sequence \begin{eqnarray} \beta_0 &=& 1,\\ \beta_1 &=& 1.005,\\ \beta_2 &=& 1.00505,\\ \vdots && \vdots\\ \beta_n &=& \alpha_n\cdot 10^{-2^n-2}. \end{eqnarray} The sequence $(\beta_n)$ is monotonic and upper bounded, and thus convergent in $\Bbb R$, and so is the sequence $(\beta_n^2)$.
Let
$$(\beta_n^2) \to \xi.$$
Clearly $\xi$ is a non-terminating decimal. Thus, as shown in the answer to this question, $f(x)$ is continuous in $\xi$, and so is $$ h(x) = f(x) - \sqrt x.$$
We therefore must have
$$\left(h\left(\beta^2_n\right)\right)\to h(\xi).$$
Since, by self-similarity of $h$, for each $n$ $$h\left(\beta_n^2\right) = 0,$$ it must be $$h(\xi) = 0,$$ that is $$f(\xi) = \sqrt \xi.$$
A little update