$T$ linear. $T$ is bounded if, and only if, $int(T^{-1}(\overline{B_{N_2}}(0,1))) \neq \emptyset$R $T: N_1 \rightarrow N_2$ linear. 
Let $A = \overline{B_{N_2}}(0,1)$. As $T$ is bounded, $T^{-1}(A)$ is a closed set in $N_1$, so $T^{-1}(A)$ is complete. Moreover, $T^{-1}(A)$ is a Baire space, since every complete metric space is a Baire space. Write $A = \cup_{n=1}^{+\infty}\overline{B_{N_2}}(0, \frac{1}{n})$. Then $T^{-1}(A) = \cup_{n=1}^{+\infty}T^{-1}\Big(\overline{B_{N_2}}(0, \frac{1}{n})\Big)$ and as $T^{-1}(A)$ is a Baire space, there is $k_0$ such that $int\big(T^{-1}(\overline{B_{N_2}}(0, \frac{1}{k_0}))\big) \neq \emptyset$.
I'm having some trouble showing that if $int(T^{-1}(\overline{B_{N_2}}(0,1))) \neq \emptyset$ then $T$ is bounded. 
My attempt: 
If $int(T^{-1}(\overline{B_{N_2}}(0,1))) \neq \emptyset$, there is $x_0 \in int(T^{-1}(\overline{B_{N_2}}(0,1)))$ and $r>0$ such that $B(x_0, r) \subset int(T^{-1}(\overline{B_{N_2}}(0,1)))$.
Then, $T(B(x_0, r)) \subset \overline{B_{N_2}}(0,1)$ 
$\implies \{T(x_0)\} + T(B(0, r)) \subset \overline{B_{N_2}}(0,1)$
$\implies \{T(x_0)\} + rT(B(0, 1)) \subset \overline{B_{N_2}}(0,1)$
$\implies  T(B(0, 1)) \subset \overline{B_{N_2}}(0,\frac{1}{r}) + \{T(\frac{-x_0}{r})\}$
It follows that $T$ is bounded, but I'm not sure about some steps...
I would be very glad if someone could help me!
 A: I only disagree with the first part of your proof. First of all, you don't know whether $N_1$ is Banach, so you can't use Baire's theorem. Additionally, when you write $T^{-1}(B_{N_2})=\cup_{n=1}^\infty T^{-1}(B_{N_2}[0,\tfrac{1}{n}])$, $T^{-1}(B_{N_2})$ already appears in this union for $n=1$. So what you deduced is that $T^{-1}(B_{N_2})$ has a relative non empty interior with respect to itself, but this is obviously true  in every metric space. 
Notice that this cannot be fixed by taking a "better" choice of sets when you take this countable union, for the exact same reason: The set, which Baire's theorem assures that has non empty interior, will have non empty interior relative to $T^{-1}(B_{N_2})$ and not necessarily relative to $N_1$.
So, I suggest a more direct approach: If $T$ is bounded, there exists an $M>0$ such that $\|Tx\|\leq M\|x\|$, for every $x\in N_1$. So $B(0,\tfrac{1}{2M})\subseteq T^{-1}(B_{N_2})$: Indeed, if $x$ is such that $\|x\|\leq \tfrac{1}{2M}$, then $\|Tx\|\leq \tfrac{1}{2}$, so $Tx\in B_{N_2}$. This implies that $0 \in   [T^{-1}(B_{N_2})]^\circ \neq \emptyset$. Notice that we didn't use any assumption regarding $N_1$ being a Banach space.
The other direction looks good the way you wrote it. Which step are you unsure about?
