Suppose $h:M \rightarrow E$ is a homeomorphism onto its image . Show that $h(M)$ is a $G_{\delta}$-set. Let $M$ and $E$ be complete metric spaces. 
    Suppose $h:M \rightarrow E$ is a homeomorphism onto its image (i.e. $h$ is a continuous one-to-one map, and $h^{-1}|_{h(M)}$ is continuous). 
    Show that $h(M)$ is a $G_{\delta}$-set.
My attempt: I think it has something to do with 
$$h(M) = \bigcap_{n \in \mathbb{N}}h(?)$$
where $?$ is open in $E.$ 
However, I do not know what precisly is $?$.

Let $B_E = \{ e \in E: \| e \| \leq 1 \}$ be the unit ball centered at $e$ with radius $1.$
Let $(\varepsilon_n)_{n \in \mathbb{N}}$ be a sequence of positive real numbers such that $\varepsilon_n \rightarrow 0$ as $n \rightarrow \infty.$
We claim that 
$$h(M) = \bigcap_{n \in \mathbb{N}}\bigcup_{y \in h(M)}(y+ \varepsilon_n \cdot B_E).$$
For each $n \in \mathbb{N},$ clearly we have $h(M) \subseteq \cup_{y \in h(M)}(y + \varepsilon_n \cdot B_E).$
Therefore, $h(M) \subseteq \bigcap_{n \in \mathbb{N}}\bigcup_{y \in h(M)}(y+ \varepsilon_n \cdot B_E).$ 
I constructed the following proof based on Batominovski's comment. 
To show another inclusion, we let $z \in \bigcap_{n \in \mathbb{N}}\bigcup_{y \in h(M)}(y+ \varepsilon_n \cdot B_E).$
By definition, for each $n \in \mathbb{N},$ there exists $y \in h(M)$ such that $z \in y+\varepsilon_n \cdot B_E.$
Therefore, as $n \rightarrow \infty,$ there exists $y \in h(M)$ such that $z = y \in h(M).$
Hence, the reverse inclusion holds. 
Since $\bigcup_{y \in h(M)}(y+ \varepsilon_n \cdot B_E)$ is open in $E,$ we conclude that $h(M)$ is a $G_{\delta}$-set. 
