Let $(X, {\lVert\; \rVert}_X),(Y, {\lVert\; \rVert}_Y)$ normed spaces and $T:X\longrightarrow Y$ a linear map. Then are equivalent:
1-$T$ is open
2-$\exists\delta >0$ such that $\{y\in Y : \lVert y\rVert_Y<\delta\}\subset\{T(x) : \lVert x\rVert_X<1\}$
3-$\exists M>0$ such that $\forall y\in Y \exists x\in X$ such that $T(x)=y$ and $\lVert x\rVert_X\leq M\lVert y\rVert_Y$
I guess $1\Longrightarrow 2$ it's clear since T is open, so let $U=\{x : \lVert x\rVert_X<1\}$ which is open in $X$, the $T(U)=\{T(x) : \lVert x\rVert_X<1\}$ is open, and since is open exists $B(0,\delta_Y)$ such that $B(0,\delta_Y)\subset\{T(x) : \lVert x\rVert_X<1\}$
I have several problems with $2\Longrightarrow 3$ and $3\Longrightarrow 1$.
For $(2)\Longrightarrow (3)$, let $y\in Y$. Define the vector $$ y'=\frac{\delta}{2\lVert y\rVert}y, $$ so that $\lVert y'\rVert=\delta/2<\delta$. Then, by (2), there exists an $x'$ with norm less than 1 such that $T(x')=y'$. Multiplying both sides by $2\lVert y\rVert/\delta$ and using the linearity of $T$, we get $$ y=T\left(\frac{2\lVert y\rVert}{\delta}x'\right), $$ and so defining $x={2\lVert y\rVert}x'/\delta$, we've got an $x$ that maps to $y$ under $T$. Finally, $$ \lVert x\rVert = \left\lVert \frac{2\lVert y\rVert}{\delta}x' \right\rVert =\frac{2\lVert y\rVert}{\delta}\left\lVert x' \right\rVert < \frac{2\lVert y\rVert}{\delta}, $$ where the last inequality holds because the norm of $x'$ is less than 1. Calling $M=2/\delta$, we're done.
For the next part of the proof, we use (3) to prove a stronger version of (2).
By (3), for every $y$ there exists an $x$ with $$ \lVert x \rVert < M \lVert y \rVert, $$ satisfying $T(x)=y$. Taking $y$ such that $\lVert y \rVert < \delta = \frac{\epsilon}{M}$, where $\epsilon>0$ an integer, this inequality becomes $$ \lVert x \rVert < M \lVert y \rVert < M\frac{\epsilon}{M} =\epsilon. $$ Thus, such a $y$ is necessarily in the set $\{T(x): \lVert x \rVert < \epsilon\}$, and hence there exists a $\delta$ (by construction) such that $$ B_Y\left(0,\delta\right) \subset \left\{T(x): \lVert x \rVert < \epsilon\right\}. $$ Note that taking $\epsilon=1$ yields (2), and hence $(3)\Longrightarrow(2)$. For the next part of the proof, note that $$ \{T(x): \lVert x \rVert < \epsilon\} = T(B_X(0,\epsilon)), $$ i.e., it is the image of the open ball of radius $\epsilon$ centered on zero.
Finally, we use the stronger version of (2) to prove (1).
Let $U\subset X$ be open. Consider $y\in T(U)$. Our goal is to construct an open ball centered on $y$ that is completely contained $T(U)$, which will show that it is open.
Since $y\in T(U)$, there is an $x\in U$ such that $T(x)=y$. Since $U$ is open, there exists an $\epsilon$ such that $B(x,\epsilon)\subset U$. The set $B(x,\epsilon)-x$ is the open ball $B(0,\epsilon)$ of radius $\epsilon$ centered on 0.
By the stronger version of (2), there exists a $\delta$ such that $B_Y(0,\delta)\subset T(B(0,\epsilon))$. Then $$ B_Y(y,\delta) = B_Y(0,\delta)+y\subset T(B(0,\epsilon)) + y. $$ Furthermore, recalling that $T(x)=y$, and using the linearity of $T$, this becomes $$ B_Y(y,\delta) \subset T(B(0,\epsilon)) + T(x) = T(B(x,\epsilon))\subset T(U), $$ since $B(x,\epsilon) \subset U$.