= Solution
If a <linear map> satisfies $\|Tx\|\le M\|x\|$, then $\|Tx-Tx'\|\le M\|x-x'\|$, so it is <continuous>. Conversely <continuity> at zero gives $\delta>0$ such that $\|x\|<\delta$ implies $\|Tx\|<1$. Apply this to $\delta x/(2\|x\|)$ for $x\ne0$ to obtain $\|Tx\|\le2\|x\|/\delta$. Hence \b[a <linear map> is <continuous> exactly when it is bounded].
For the separately <continuous> <bilinear map>, put $T_x(y)=F(x,y)$ for $\|x\|\le1$. Every $T_x:Y\to Z$ is a bounded <linear map> by the first result. For each fixed $y$, <continuity> in the other variable gives $\|F(x,y)\|\le C_y\|x\|$, so $\sup_{\|x\|\le1}\|T_xy\|<\infty$. The <Uniform boundedness principle> therefore gives $\sup_{\|x\|\le1}\|T_x\|=M<\infty$, and bilinearity then yields
$$
\boxed{\|F(x,y)\|\le M\|x\|\|y\|}.
$$
To spell out the uniform-boundedness step, define the closed subsets $E_m=\{y:\sup_{\|x\|\le1}\|T_xy\|\le m\}$ of the <Banach space> $Y$. They cover $Y$. The <Baire category theorem> gives an $E_m$ containing a ball $B(y_0,r)$. For $\|h\|<r$, both $y_0+h$ and $y_0$ lie in $E_m$, so $\sup_x\|T_xh\|\le2m$. Scaling $h=(r/2)y$ for $\|y\|\le1$ bounds all the operator <norms> by $4m/r$. This proves the claimed uniform estimate without presuming joint <continuity>.
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