Prove the boundedness of a bilinear continuous mapping. Let $X,Y,Z$ are Banach spaces and $$B:X\times Y\to Z$$ is bilinear and continuous. Prove that there exists $M<\infty$ such that 
$$\lVert B(x,y)\rVert \leq M\lVert x\rVert\lVert y\rVert.$$ Is completeness needed here? The bilinearity of $B$ means for a fixed  $x\in X$,$B_x\colon Y\to Z$ is linear and for a fix $y\in Y$, $B^y:X\to Z$ is linear.
Since $X,Y,Z$ are metrizable by norms, I think what we do is to show that $B$ is bounded. Let $E_1,E_2$ are bounded, then $E_1\subset t_1V_1$ and $E_2\subset t_2V_2$ for all large $t_1,t_2$; put $M=\max\{t_1,t_2\}$, so that $$B(E_1\times E_2)\subset B(t_1V_1\times t_2V_2)\subset MB(V_1\times V_2)\subset MW,$$ in which $V_1,V_2$ are balanced neighborhoods of  origins in $X,Y$ respectively and $W$ is a neighborhood of $0$ in $Z$.
Can someone point out if there is some errors in my thought or give a straightforward path?
 A: Although Giuseppe mentioned in his comment, I am giving the little details of the alternative proof.Note that for $x\in X_1$ (the unit ball of $X$), the collection {${B_x| x\in X_1}$} is a set of continuous linear map (By continuity of $B$ in 2nd variable) from the Banach Space $Y$ to $Z$. And the collection is also pointwise bounded.(By contunuity of $B$ in 1st variable).Hence by uniform boundedness principle we will have {${B_x| x\in X_1}$} is uniformly norm bounded. And hence the required result follows. 
A: This post is just a supplement and detailed version of Timon's answer.
I think we may need to change $X_1$ to unit circle. So in this post, $X_1$ will be the unit circle in $X$. And the rest of notations are the same with those in Timon's answer.
Following from Timon's answer, we know that, $\forall x\in X_1,\,y\in Y$, we will have that $|B(x,y)|\le||B_x|| \cdot||y|| \le M \cdot ||y||$, where $M:=\sup_{x\in X_1} ||B_x||<\infty$.
Note that $\forall x\in X_1$, $||x||=1$.
Hence, $\forall x\in X_1,\,y\in Y$, we have that $|B(x,y)|\le M \cdot ||y||\cdot ||x||$
Now for any arbitrary $\alpha\in X,$ we know there exists an $x_0 \in X_1$ and some constant $c$ s.t. $\alpha=c\, x_0.$
Therefore, $$|B(\alpha,y)|=|B(c\,x_0,y)|=|cB(\,x_0,y)|=|c|\cdot|B(\,x_0,y)|\, \le |c|\cdot M \cdot ||y||\cdot ||x_0||$$
$$=M \cdot ||y||\cdot ||c\,x_0||= M \cdot ||y||\cdot ||\alpha||$$
I.e. $|B(\alpha,y)|\le M \cdot ||y||\cdot ||\alpha||,\,\forall \alpha\in X,\,y\in Y$.
$\tag*{$\blacksquare$}$
