Is't a correct observation that No norm on $B[0,1]$ can be found to make $C[0,1]$ open in it? There's a problem in my text which reads as:

Show that $C[0,1]$ is not an open subset of $(B[0,1],\|.\|_\infty).$

I've already shown in a previous example that for any open subspace $Y$ of a normed linear space $(X,\|.\|),~Y=X.$ Even though using this result this problem turns out to be immediate the sup-norm is becoming immaterial. 
And I can't believe what I'm left with:
No norm on $B[0,1]$ can be found to make $C[0,1]$ open in it.

Is this a correct observation?
 A: This statement is actually true under more general settings. It seems convenient to talk about topological vector spaces, of which normed spaces are a very special kind.
So let $X$ be a topological vector space and $Y$ be an open subspace. So we know $Y$ contains some open set. 
Since the topology on topological vector spaces are translation invariant (that is, $V$ is open if and only if $V+x$ is open for all $x\in X$, you can check this in normed spaces), we know $Y$ contains some open neighborhood of the origin, say $0\in V\subset Y$.
Another interesting fact about topological vector spaces is that for any open neighborhood $W$ of the origin, one has \begin{equation}
X=\cup_{n=1}^{\infty}nW.
\end{equation}Again you might check this for normed spaces. Apply this to our $V$, and note that $Y$ is closed under scalar multiplication, we have \begin{equation}
X=\cup nV\subset \cup nY=Y.
\end{equation} 
So we have just proved 

The only open subspace is the entire space.

Note: If $S$ is a subset of a vector space, for a point $x$ and a scalar $\alpha$ we define \begin{equation}
x+S:=\{x+s|s\in S\}
\end{equation} and \begin{equation}
\alpha S:=\{\alpha s|s\in S\}.
\end{equation} 
A: This is true.  Indeed, you can show more:

Let $X$ be a normed vector space.  If $E \subseteq X$ is a linear subspace which is open, then $E = X$.

My guess is this follows from the same argument you already have.
