Different textbooks give different, but equivalent, formulations of the open mapping theorem, and I want to understand how all of these rigorously relate (and how they're all equivalent). I intuitively understand why they're equivalent, but want a deeper formal understanding. So, here are the various formulations:
Rudin: Let $X,Y$ be Banach spaces, and $T \in \mathcal{L}(X,Y)$ (set of all bounded linear maps from $X$ to $Y$) be a surjective map. Then $T$ is an open map.
Note that Rudin's formulation doesn't include the converse of this statement, although it can easily be checked.
Royden: Let $X,Y$ be Banach Spaces and $T \in \mathcal{L}(X,Y)$. Then $\text{Im}(T)$ is closed if and only if the operator $T$ is open.
Brezis: Let $X,Y$ be Banach Spaces and $T \in \mathcal{L}(X,Y)$. Then there is $r > 0$ such that $B_{r}^{Y}(0) \subset T(B^{X}_{1}(0)$), where $B^{E}_{r}(x)$ denotes the open ball of radius $r$ about $x$ in space $E$.
Intuitively, I can see how these formulations are equivalent. If the image if closed and the operator is open, the image is an open and closed space in $Y$, which means that it's $Y$ itself (i.e. the map is surjective). Similarly, in the third formulation, if the image of unit ball contains a neighborhood of the origin, then by linearity all open sets will be mapped to open sets. I do, however, struggle with formalizing these ideas, and would appreciate if someone could (fairly rigorously) explain the equivalence of these notions.
First, and as stated in the comments, observe that both Royden and Brezis statements are false (at least as you stated them). For a counterexample take $T=0$ and $X=Y=\mathbb{R}$. With this being said, I think we can rescue some of the main ideas:
Royden's statement is actually true if you say that an operator is open when it takes open sets into open sets of the induced topology of $Y$ into $\text{Im}(T)$, that is, for every open set $O\subset X$, $T(X)=O'\cap \text{Im}(T)$, where $O'$ is an open set of $Y$. In such case, Rudin and Royden statements are exactly the same considering $\text{Im}(T)=Y$, observing that in both cases the maps are onto operators between banach spaces.
In order to connect the first two statements with the last one it's necessary to state that we need surjectivity as an assumption on Brezis's statement. In such case it would be enough to prove that a map is open if and only if $B_r^Y(0)\subset B_1^X(0)$ for some $r>0$, so we'll prove that:
($\Rightarrow$) If $T$ is open it maps $B_1^X(0)$ into a open set of $O\subset Y$. Since $T$ is linear $T(0)=0$, and so $o\in O$. Finally, since $O$ is open, there exist $r>0$ such that $B_r^Y(0)\subset O=T(B_1^X(0))$.
($\Leftarrow$) Let $r>0$ such that $B_r^Y(0)\subset T(B_1^X(0))$ and $O\subset X$ an open set. We'll prove that $T(O)$ is open on $Y$, for that let $y\in T(O)$ and note that by its definition there exist $x\in O$ such that $T(x)=y$. Since $O$ is open there is $\lambda>0$ such that $B_\lambda^X(x)\subset O$, and so using the linearity of $T$ on $B_r^Y(0)\subset T(B_1^X(0))$ we have that $$ \begin{align} \lambda B_r^Y(0)&\subset \lambda T(B_1^X(0))\\ B_{\lambda r}^Y(0)&\subset T(\lambda B_1^X(0))\\ B_{\lambda r}^Y(0)&\subset T(B_\lambda^X(0))\\ B_{\lambda r}^Y(0)+y&\subset T(B_\lambda^X(0))+y\\ B_{\lambda r}^Y(y)&\subset T(B_\lambda^X(0))+T(x)\\ B_{\lambda r}^Y(y)&\subset T(B_\lambda^X(x))\subset T(O),\\ \end{align} $$ which proves that $T(O)$ is an open set on $Y$.