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.
We have no examples of unbounded linear operators between these spaces. The Axiom of Choice tells us there are many such operators, but it does not tell us what they are, so we don't have any way to write down such an operator.
Wikipedia goes into this subject too:
As noted above, the axiom of choice (AC) is used in the general existence theorem of discontinuous linear maps. In fact, there are no constructive examples of discontinuous linear maps with complete domain (for example, Banach spaces).
[...] it is consistent with set theory without AC that there are no discontinuous linear maps on complete spaces. In particular, no concrete construction such as the derivative can succeed in defining a discontinuous linear map everywhere on a complete space.
With AC, the proof of existence of an unbounded linear map $T:X\to Y$, for any $X, Y$, can proceed like this: let $\{v_\alpha\}$ be a Hamel basis of $X$ and pick any nonzero vector $w\in Y$. Take any unbounded collection of numbers $\{c_\alpha\}$ indexed by the same set as $\{v_\alpha\}$. Define $T(v_\alpha) = c_\alpha \|v_\alpha\| w$ and extend by linearity (which is possible since $\{v_\alpha\}$ is a basis).
The above does not produce an example of an unbounded linear operator on an infinite-dimensional Banach space because there are no examples of a Hamel basis for such a space.
Best Answer
Why does this confuse you? As you proved, $T_n\to 0$ in norm. Also, it is true that $T_n(x)\to 0$ for all $x\in\mathbb{R}$. What you are looking at, $T_n(n)=1$, is true but this is irrelevant. When we say $T_n(x)\to 0$ for all $x$, $x$ is fixed and $n$ tends to infinity. In the expression $T_n(n)$ the argument is not fixed, so it is not relevant to the rest of our observations.
It might help you think that if we fix $m\in\mathbb{N}$, then $T_n(m)\to0$ as $n\to\infty$.