Let $E$ be metrizable space, that is, there exists a basis $\mathcal{B}:=\{U_n \subset E \; ; \; n\in \mathbb{N}\}$ of neighborhoods of $0 \in E$. I want to prove that: if $\{B_n \subset E \; ; \; n \in \mathbb{N}\}$ is family of bounded sets, then there exists a sequence $(\varepsilon_n)_{n \in \mathbb{N}} \subset \mathbb{R}_+^{*}$ such that $$B:= \bigcup_{n=0}^{\infty}\varepsilon_n B_n$$ is a bounded subset of $E$. I thought of the following: since, for each $n \in \mathbb{N}$, $B_n \subset E$ is bounded, and $U_n$ is a neighborhood of $0 \in E$, then exist $\lambda_n>0$ satisfying $$B_n \subset \lambda_n U_n.$$ Define, for each $n \in \mathbb{N}$, $\varepsilon_n:=\frac{1}{\lambda_n}$. So, $$B=\bigcup_{n=0}^{\infty}\varepsilon_nB_n \subset \bigcup_{n=0}^{\infty}U_n.$$
For this, how do I conclude that given a neighborhood of $ 0 \in E $ there is $ \lambda> 0 $ such that $ B \subset \lambda V $?
Let $V_1 \supset V_2 \supset \cdots \supset V_m \supset \cdots$ be a countable decreasing basis of neighborhoods of $0$ in $E$ where $V_i$ is balanced for all $i\in \mathbb{N}$. For each $k \in \mathbb{N}$ select $\varepsilon>0$ such that $\varepsilon_k B_k \subset V_k$. We claim that the sequence $\{\varepsilon_k\}$ fulfills the desired properties. Indeed, let $n \in \mathbb{N}$ arbitrary. If $k\geq n$ then $\varepsilon_k B_k \subset V_k \subset V_n$. There is $c_n>0$ such that $c_n \varepsilon_k B_k \subset V_n$ for $k<n$, since $\bigcup_{k=1}^{n}\varepsilon_kB_k$ is bounded. Thus, $B \subset \frac{1}{c_n}V_n \cup V_n$. Let $\lambda_n=\max\{1,\frac{1}{c_n}\}$ then $B \subset \lambda_n V_n$. Therefore, $B$ is bounded.