$f$ is open iff the image of the unit ball contains a ball centered at 0 Disclaimer: I asked a similar question earlier but it was poorly phrased and not specific enough. Upon editing that question I realized it was too different compared to the original, so I have deleted that one and have posted this one instead.
In Folland's Real Analysis he states:

If $X$ and $Y$ are topological spaces, a map $f: X \rightarrow Y$ is called open if $f(U)$ is open in $Y$ whenever $U$ is open in $X$. If $X$ and $Y$ are metric spaces, this amounts to requiring that if $B$ is a ball centered at $x \in X$, then $f(B)$ contains a ball centered at $f(x)$. Specializing still further, if $X$ and $Y$ are normed linear spaces and $f$ is linear, then $f$ commutes with translations and dilations; it follows that $f$ is open iff $f(B)$ contains a ball centered at $0$ in $Y$ when $B$ is the ball of radius $1$ about $0$ in $X$.

To show that $f(B)$ is open we need to show that for all points $p \in f(B)$ there exists a ball $B(p,r) \subset f(B)$ centered at $p$ and with sufficiently small radius $r$. I think the idea here is that if we can find a ball about $0$ in $Y$, then we can find open neighborhoods at all points $p \in f(B)$ by translating and dilating so that the ball contains $p$ but is still contained in $f(B)$.
However  since there is no continuity assumed on $f$, how can we ensure that this can always be done? For example, suppose $f$ maps the unit ball in $X$, $B_X$ to the unit ball in $Y$ minus one point $B_Y - \{y\}$, where $y \neq 0$ and we define $y = f(x')$ to be outside $B_Y$. If $\|y\| = r$, we can choose any ball centered about $0$ in $Y$ with radius less than $r$, then we have constructed a map $f$ such that $f(B_x)$ contains a ball centered at $0$ in $Y$, and hence by the above definition should be open. However, $f(B_X)$ is not open by the construction outlined above.
 A: You say

However since there is no continuity assumed on $f$, how can we ensure that this can always be done?

The hypothesis that $0$ is an interior point of $f(B_X)$ is precisely equivalent to  $f$ being open. There is no direct relationship with continuity; for instance every linear functional is open, even if not continuous.

A linear map cannot map $B_X$ to $B_Y\setminus\{y\}$. Because there is some $v\in B_Y$ small enough such that $y\pm v\in B_Y$. Then there would exist $w,z\in B_X$ such that $f(w)=y+v$ and $f(z)=y-v$. And then
$$
f\big(\frac12(w+z))=\frac12\,(f(w)+f(z))=y.
$$

In any case, the argument to show that if $f$ is linear and $0$ is an interior point of $f(B_X)$, then $f$ is open, is fairly straightforward. As Folland says, it's all due to the fact that $f$ commutes with translations and dilations.  Indeed, fix $U\subset X$ open, and let $y\in f(U)$; we want to show that $y$ is an interior point in $f(U)$.
Notation: $B_r^X(z)$ is the ball in $X$ of radius $r$ and centre $z$. In particular $B_X$ above is $B_1^X(0)$.
Choose $x\in U$ such that $f(x)=y$. Let $U'=-x+U$. Then $U'$ is open and $0\in U'$. So there exists $\delta>0$ such that $B_\delta^X(0)\subset U'$. By hypothesis there exists $r>0$ such that $B_r^Y(0)\subset f(B_1^X(0))$. As $f$ is linear, this means that
$$
B_{\delta r}^Y(0)\subset \delta f(B_1^X(0))=f(\delta B_1^X(0))=f(B_\delta^X(0))\subset f(U')=-f(x)+f(U).
$$
Then
$$
B_{\delta r}^Y(f(x))=f(x)+B_{\delta r}^Y(0)\subset f(x)-f(x)+f(U)=f(U). 
$$
So we have shown that  any $y\in f(U)$, there exists $\delta'>0$ such that $B_{\delta'}^Y(y)\subset f(U)$; that is, $f(U)$ is open.
