To expand on my comment:$\DeclareMathOperator{\Hom}{Hom}$
There are many ways to equip $X \times Y$ with a norm, the most natural ones are $\|(x,y)\| = \max{\{\|x\|,\|y\|\}}$ and $\|(x,y)\| = \|x\| + \|y\|$ since they correspond to the categorical product and coproduct operations (in the category of Banach or normed linear spaces and linear maps of norm $\leq 1$). Be that as it may, it is a good exercise to check that all the $p$-norms $\|(x,y)\|_{p} = \left(\|x\|^{p} + \|y\|^{p}\right)^{1/p}$ on $X \times Y$ are equivalent, thus a linear map $T: X \times Y \to Z$ is continuous if and only if it is bounded with respect to any of the norms with which you can equip the space $X \times Y$.
Note that linear means $T(\lambda x, \lambda y) = \lambda T(x,y)$ and $T(x+x',y+y') = T(x,y) + T(x',y')$ for all $(x,y), (x',y') \in X \times Y$ and all $\lambda \in \mathbb{R}$.
The multiplication is not linear in this sense, but it is bilinear $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$. As such, it is obviously continuous and it is bounded with respect to the norm on the bilinear maps $B: X \times Y \to Z$ given by
$$\|B\| = \sup_{\|x\|,\|y\| \leq 1} \|B(x,y)\|_{Z}$$
and it is a good exercise to check:
- A bilinear map $B$ is continuous if and only if it is bounded with respect to the norm above. Note that this simply means that $\|B(x,y)\|_{Z} \leq \|B\| \,\|x\|_{X} \, \|y\|_{Z}$ by bilinearity.
- If $Z$ is a Banach space then the space of bilinear maps $X \times Y \to Z$ is complete with respect to that norm.
Added: In fact, this idea naturally leads to the notion of the projective tensor norm.
Recall that a bilinear map $B: X \times Y \to Z$ corresponds to a linear map $b: X \otimes Y \to Z$. An element of the tensor product can be written as a finite sum $\sum x_i \otimes y_i$ and since $b$ is linear we have $b(\sum x_i \otimes y_i) = \sum b(x_i \otimes y_i) = \sum B(x_i,y_i)$. Now we want a norm on $X \otimes Y$ such that $b$ is bounded if and only if $B$ is bounded. Now $B$ is bounded if and only if $B$ is bounded on the elementary tensors, for which we have $\|B(x,y)\| \leq \|B\|\,\|x\|\,\|y\|$ so the norm of $\sum x_i \otimes y_i$ on $X \otimes Y$ had better be comparable to $\sum \|x_i\| \, \|y_i\|$. Now the problem is that this is not well defined because an element of $X \otimes Y$ has many decompositions into sums of elementary tensors. It turns out that the correct definition for the norm of $\omega \in X \otimes Y$ is
$$\|\omega \|_{\pi} = \inf\left\{ \sum \|x_i\|_X \, \|y_i\|_{Y} \,:\, \omega = \sum x_i \otimes y_{i}\right\}$$
where the infimum is taken over all (finite) representations $\omega = \sum x_i \otimes y_i$ as sum of elementary tensors. It is quite obvious that $\|\cdot\|_{\pi}$ is a semi-norm on $X \otimes Y$ satisfying $\|x \otimes y\|_{\pi} \leq \|x\|_{X} \|y\|_{Y}$. A bit more work shows that actually $\|x \otimes y\|_{\pi} = \|x\|_{X} \|y\|_{Y}$ and that $\| \cdot \|_{\pi}$ is a norm. Moreover, one can check that $\|B\| = \|b\|$, when the latter is computed as operator norm on $X \otimes Y$ with respect to $\|\cdot \|_{\pi}$. This gives us a bijection
$$\text{Bil} (X,Y;Z) = \text{Hom}(X \otimes Y, Z)$$
between the spaces of bounded bilinear maps $X \times Y \to Z$ and bounded linear maps $X \otimes Y \to Z$. Now if $X,Y$ happen to be Banach spaces then $X \otimes Y$ no longer is a Banach space in general, so we may simply complete it and we write $X \widehat{\otimes} Y$ for this completion. Combining this with the observation made by Mark in his answer, we get the (isometric) correspondences
$$\Hom{(X \widehat{\otimes} Y, Z)} = \text{Bil}(X,Y;Z) = \Hom(X,\Hom(Y,Z))$$
which we know well from linear algebra. We equip all $\Hom$-spaces with their natural operator norms and the space $\text{Bil}$ with the norm I defined above.
Best Answer
We will first reduce the problem to a nonzero bilinear map from $\mathbb{R}^2$ to $\mathbb{R}$.
Take any $x \in E$ and $y \in F$ such that $T(x,y) = z$ for some $0 \neq z \in G$ and consider the subspaces $A = \operatorname{span} \{x\} \times \operatorname{span} \{y\}$ of $E \times F$, and $B = \operatorname{span} z$ of $G$. Now just identify $A$ with $\mathbb{R}^2$ and $B$ with $\mathbb{R}$ and consider the restriction of $T$ on $A$ with range in $B$. If we show that this restriction is not uniformly continuous, then so cannot be the original map $T$.
So it remains to show that any nonzero bilinear map $T$ from $\mathbb{R}^2$ to $\mathbb{R}$ is not uniformly continuous. Note that any such map is of the form $T(x,y) = c \cdot xy$ for some nonzero constant $c$ (just take $c = T(1,1)$). It is now an easy excercise to show that this $T$ is not unifromly continuous.