Let $V,W$ be any finite dimensional normed real vector spaces, and $X$ any normed real vector space. Let $B:V\times W \to X$ be a bilinera map. I want to show that $B$ is continuous.
My argument
Since on a finite dimensional real vector space all norms are equivalent, I can choose to consider the following norms on $V,W$ and $V\times W$ : choose a basis $(E_1,\dots, E_n)$ of $V$ and $(F_1,\dots,F_m)$ of $W$ and define $$|\cdot|_V:V\to [0,+\infty), v=\sum_i v^iE_i\mapsto\sum_i|v_i|$$ and $$|\cdot|_W:W\to [0,+\infty), v=\sum_i v^iF_i\mapsto\sum_i|v_i|$$, and $$|\cdot|:V\times W\to [0,+\infty), (v,w)\mapsto |v|_V+|w|_W$$
Now we have $B(v,w)=\sum_{i,j}v^iw^jB(E_i,F_j)$ and putting $M=\max\{|B(E_i,F_j)|_X:i,j\}$ we have
$$|B(v,w|\leq M\sum_{i,j}|v^iw^j|\leq Mnm |v||w|$$ and from this it follows
$|B(v,w)-B(h,k)|=|B(v,w)-B(v,k)+B(v,k)-B(h,k)|\leq|B(v,w-k)|+|B(v-h,k)|\leq Mmn[|v||w-k|+|v-h||k-w|+|v-h||w|]\leq Mmn [|v|+|(v,w)-(h,k)|+|w|]|(v,w)-(h,k)|$.
Is my argument correct? I need this for showing that $B$ is differentiable according to the definition of Lee's "Introduction to smooth manifolds " 2° Edition Appendix C pag 642.
In my proof of differentiability I need continuity so I'm proving this first.
If you know shorter ways of showing continuity of $B$ let me know. Also let me know if you can prove differentiability of $B$ without first showing continuity.
N.B. I can only use definitions and results previously showed in that book.
Your arguments for continuity are correct.
If you know about tensor product, you can make an alternative proof:
Observe that $B:V\times W\to X$ is just the composition of the following maps: $$V\times W\ \overset{\vartheta}\to\ V\otimes W\ \overset{\hat B}\to\ S\ \hookrightarrow\ X$$ where $\vartheta$ is the canonical bilinear map $(v,w)\mapsto v\otimes w$; $\ \hat B$ is the linear map induced by $B$, mapping $v\otimes w\mapsto B(v,w)$, and $S:={\rm ran}(\hat B)$ is its image, which is included in $X$.
Now, all these, except perhaps $X$, are finite dimensional real vector spaces, so both $\vartheta$, as a bilinear map, and $\hat B$, as a linear map, are differentiable (even smooth).
So is the inclusion of $S$ into $X$, as that's also a bounded linear map.
Edit: Actually, we can even skip the tensor product part, and define $S$ as the range of the (bilinear) function $B$, and show that it's a finite dimensional subspace (its dimension is $\le\dim V\cdot\dim W$).