Let $X$ be a topological space. Assume that there exists a sequence of simple functions $\phi_n:X\to X$ (finite range and measurable) with $\lim\phi_n(x)=x$. Can we concluded $X$ may be written by a countable union $X=\cup X_n$ where all $X_n$'s are all relatively second countable?

2$\begingroup$ By measurable do you mean Borel measurable, or is it something else? $\endgroup$ – Andrés E. Caicedo May 18 '18 at 20:44

$\begingroup$ @ Andrés E. Caicedo : Yes, I mean Borel measurable. $\endgroup$ – Ali Bagheri May 18 '18 at 20:48
The answer is no. The Moore plane is a counterexample.
For $z \in \mathbb{R}^2$ and $r \geqslant 0$ denote by $\bar D(z, r)$ the closed Euclidean disc of center $z$ and radius $r$. Recall that the Moore plane is $M = [0, + \infty) \times \mathbb{R}$, where:
 A basis of neighborhoods of $(x, y) \in (0, +\infty) \times \mathbb{R}$ consists in all discs $\bar D((x, y), r)$ with $0 < r < x$;
 A basis of neighborhoods of $(0, y)$ consists in all discs $\bar D((r, y), r)$, with $r>0$.
This space is a usual example of a separable topological space that is not secondcountable. To see that $M$ is separable, remark that the set $D = \{(x, y) \in \mathbb{Q}^2 \mid x> 0\}$ is dense in $M$. However $M$ cannot be written as a countable union of secondcountable spaces, otherwise the subspace $\{0\} \times\mathbb{R}$ could also be written as such a union; but the inherited topolgy on $\{0\} \times\mathbb{R}$ is discrete.
To show that an approximation of the identity as required exists in $M$, let $(r_k)_{k \in \mathbb{N}}$ be an enumeration of $D$ and define $\phi_n$ as follows: for $z \in M$, consider $r \geqslant 0$ minimal such that the basic neighborhood of $z$ of radius $r$ contains an element of $\{r_0, \ldots r_n\}$, and let $\phi_n(z) = r_k$ be such an element with $k$ minimal. Then $\phi_n(z) \rightarrow z$: indeed, if $(r_{k_i})$ is a sequence of elements of $D$ converging to $z$, then for $U$ a basic neighborhood of $z$, there is $i_0 \in \mathbb{N}$ such that $r_{k_{i_0}} \in U$; and then, for all $n \geqslant k_{i_0}$, we have $\phi_n(z) \in U$. Obviously, $\phi_n$ has finite range, included in $\{r_0, \ldots, r_n\}$. And to show that $\phi_n$ is Borel, remark that for $k \in \{0, \ldots, n\}$, $\phi_n^{1}(\{r_k\}) = A \sqcup B$, where $A \subseteq \{0\}\times \mathbb{R}$ and $$B = \{z \in (0, \infty) \times \mathbb{R} \mid (\forall 0 \leqslant i < k, \, d(z, r_k) < d(z, r_i)) \wedge (\forall k < i \leqslant n, \, d(z, r_k) \leqslant d(z, r_i))\},$$ where $d$ is the Euclidean distance in $\mathbb{R}^2$.
$A$ is relatively open in $\{0\} \times \mathbb{R}$ as the inherited topology on this subset is discrete. As $\{0\} \times \mathbb{R}$ is closed in $M$, then $A$ is Borel in $M$. In the same way, $B$ is Borel in $\mathbb{R}^2$, so since the topology inherited from $M$ on $(0, +\infty) \times \mathbb{R}$ is the same as the topology inherited from $\mathbb{R}^2$, and since $(0, +\infty) \times \mathbb{R}$ is open in $M$, we deduce that $B$ is Borel in $M$. So $\phi_n^{1}(\{r_k\})$ is Borel in $M$.

1$\begingroup$ @NdeRancourt The Moore plane (rotated at 90 degrees) is wellknown in General Topology as the Niemycki plane (see page 22 of Engelking's "General Topology"). $\endgroup$ – Taras Banakh May 19 '18 at 6:45
Here is a counterexample where the $\phi_n$'s are continuous.
Let $X$ be the Sorgenfrey line, that is $X=\mathbb{R}$ with the topology generated by intervals of the form $[a,b)$ with $a,b \in \mathbb{R}$. Note that $X$ is not a countable union of second countable subspaces since for any $Y \subseteq X$ the size of a base for $Y$ must have cardinality at least $Y$ (so that $Y$ is second countable only if $Y$ is countable).
Fixing an enumeration of the rationals $\mathbb{Q}=\{q_i:i \in \omega\}$, we can define $$\phi_n(x)=\min \{q_i:i\leq n\}\cap (x,\infty)$$ if the set $\{q_i:i\leq n\}\cap (x,\infty)$ is nonempty and $\phi_n(x)=\max \{q_i:i \leq n\}$ otherwise. It is easy to see that each $\phi_n$ is continuous and $\lim\phi_n(x)=x$ for every $x \in X$.