Prove that a linear operator $L$ between two normed spaces $(X_1,ν_1)$ and $(X_2,ν_2)$ is bounded iff $ν_2\big(L(x_1)\big)\le M⋅ν(x_1)$ with $M>0$. 
Definition
A function $f$ between two metric spaces $(X_1,\delta_1)$ and $(X_2,\delta_2)$ is bounded if it maps bounded set into bounded set.

So with respect the last definition I am trying to prove that if a linear operator $L$ between two normed spaces $(X_1,ν_1)$ and $(X_2,ν_2)$ is bounded if and only if there exists $M\in\Bbb R^+$ such that the inequality
$$
\begin{equation}\tag{1}\label{1}\nu_2\big(L(x_1)\big)\le M\cdot\nu_1(x_1)\end{equation}
$$
holds for any $x_1\in X_1$.
So on this purpose let be $Y_1\in\mathcal P(X_1)$ bounded so that there exist $y_1\in X_1$ and $\epsilon_1\in\Bbb R^+$ such that
$$
Y_1\subseteq B(y_1,\epsilon_1)
$$
Now if $\eqref{1}$ holds then by linearity we observe that for any $x_1\in X_1$ the inequality
$$
\begin{equation}\tag{2}\label{2}\nu_2\big(L(x_1)-L(y_1)\big)=\nu_2\big(L(x_1-y_1)\big)\le M\cdot\nu_1(x_1-y_1)<M\cdot\epsilon_1\end{equation}
$$
holds so that we conclude that actually the inclusion
$$
\begin{equation}\tag{3}\label{3} L[Y_1]\subseteq B\big(L(y_1),M\cdot\epsilon_1\big)\end{equation}
$$
holds and thus we can state that $L[Y_1]$ is bounded.
So as you can see I am not able to prove the converse so that I thought to put a specific question where I ask to do it and where I ask even to test the proof I gave about the first implication. So could someone help me, please?
 A: For the forward implication, consider the closed unit ball in $X_1$, $\bar B_1(0)$.  (Recall that in a general metric space, $\bar B_\varepsilon(x) = \{ y \mid d(x, y) \le \varepsilon \}$ is not necessarily the closure of $B_\varepsilon(x)$; though it happens that in the special case of a normed space, that does turn out to be true.)
Since $\bar B_1(0)$ is bounded, by hypothesis, we have that $L(\bar B_1(0))$ is bounded.  That implies that there is an upper bound $M > 0$ on norms of elements of $L(\bar B_1(0))$ (for example, if $L(\bar B_1(0)) \subseteq \bar B_\varepsilon(y))$ then $M = \varepsilon + \nu_2(y)$ will work).  We now claim that this $M$ satisfies the desired condition.  To see this, suppose $x_1 \in X_1$.  If $x_1 = 0$, then $Lx_1 = 0$, so $\nu_2(Lx_1) \le M \cdot \nu_1(x_1) = 0$.  Otherwise, $\frac{1}{\nu_1(x_1)} x_1 \in \bar B_1(0)$, so $\frac{1}{\nu_1(x_1)} \nu_2(Lx_1) \le M$, and again in this case we get that $\nu_2(Lx_1) \le M \cdot \nu_1(x_1)$.
A: Well, suppose that the above inequality does not hold. This means that you can find a sequence $(x_n)_{n \in \mathbb N}$ in $X_1$ such that
$$ \nu_1(x_n) = 1 \text{ for all } n \in \mathbb N \quad \text{and} \quad \nu_2(L x_n) \to \infty \text{ as } n \to \infty.$$
So set $A := \{x_n : n \in \mathbb N\}$. Then $A$ is obviously bounded, whereas $L[A]$ is not. Thus, $L$ is not bounded and your missing implication follows by contraposition.
Edit to answer your comment: Suppose (1) holds for all $x_1 \in X_1$. Then it trivially holds also for all $x_1 \in X_1$ with $\nu_1(x_1) = 1$.
Now suppose that (1) holds only for all $x_1 \in X_1$ with $\nu_1(x_1) = 1$ and let $x \in X_1$. If $x = 0$, then (1) clearly holds also for $x$. If $x \neq 0$, then $y := \frac{1}{\nu_1(x)} x \in X_1$ satisfies $\nu_1(y) = 1$, Thus,
$$\frac{1}{\nu_1(x)} \nu_2(Lx) = \nu_2(Ly) \leq M \nu_1(y) = M,$$
which implies (1) for all $x \in X_1$.
So (1) is equivalent to the statment that $\nu_2(Lx_1) \leq M$ for all $x_1 \in X_1$ with $\nu_1(x_1) = 1$. Now suppose that (1) does not hold, i.e., $\nu_2(Lx_1) \leq M$ does not hold for any choice of $M > 0$, then there must be a sequence with the above properties since you can find for each $n \in \mathbb N$ some $x_n \in X_1$ with $\nu_1(x_n) = 1$ such that $\nu_2(L x_n) \geq n$. Else you would find an $M > 0$ that does the job.
