For a normed space $(X, \|.\|), $ show that $(x,y) \mapsto x + y $ and $(\alpha,x) \mapsto \alpha x$ are continuous operators with respect to the norm.
Let $f_1, f_2$ be addition and scalar multiplication, respectively.
Then $$f_1 : (X, \|.\|) \times (X, \|.\|) \rightarrow (X, \|.\|)$$
Since we're trying to show continuity we have to work with their associated metric spaces:
$$f_1 : (X, \|x-y\|) \times (X, \|x-y\|) \rightarrow (X, \|x-y\|)$$
Which means that we have to show $$(\forall \epsilon \gt 0)(\exists \delta \gt 0)(d'((x,y),(x',y')) \lt \delta \Rightarrow d(f_1(x,y), f_1(x',y')) \lt \epsilon)$$ where $d'$ is the metric on the cartesian product of vector spaces $X \times X$.
This reduces to:
$$(\forall \epsilon \gt 0)(\exists \delta \gt 0)(\|(x-x', y-y')\|_2 \lt \delta \Rightarrow \|x+y - x'-y'\| \lt \epsilon)$$
Where $\|.\|_2$ means the induced metric on the cartesian product. But from here I'm having trouble understanding how to use the metric $\|.\|_2$ to show anything further.
Anyone have any ideas?
Each normed space $(X,\lVert.\rVert)$ is a topological space, its topology being the metric topology induced by $\lVert.\rVert$, i.e. by the metric $d(x,y) = \lVert x-y \rVert$. You know that continuity of maps between metric spaces can be described via the induced metric topologies (e.g. preimages of open sets are required to be open).
As you know the product of topological spaces is given the product topology. In that sense the continuity of sum $\sigma :X \times X \to X$ and scalar multiplication $\mu : \mathbb R \times X \to X$ can be verified without using metrics or norms on the products $X \times X$ and $\mathbb R \times X$. You can do these verifications explicitly if you want. For example, to verify that $\sigma$ is continuous, you can show that for each $(x,y) \in X \times X$ and each open neigborhood $U$ of $\sigma(x,y) = x+y$ in $X$ there exists an open neighorhood $V$ of $(x,y)$ in $X \times X$ such that $\sigma(V) \subset U$. Since we work with the product topology, it suffices to show that there are open neighborhoods $V_1$ if $x$ and $V_2$ of $y$ in $X$ such that $\sigma(V_1 \times V_2) \subset U$. This can be done by a standard $\varepsilon$-$\delta$-argument.
Alternatively you can easily show that the product topology on $A \times B$, where $A$ and $B$ are normed spaces with norms $\lVert.\rVert_A$ and $\lVert.\rVert_B$ respectively, is induced by a norm $\lVert.\rVert$ on $A \times B$. This norm is not unique (in contrast to the product topology). Some standard choices are
$\lVert(a,b)\rVert_2 = \sqrt{\lVert a\rVert_A^2 +\lVert b\rVert_B^2 } $
$\lVert(a,b)\rVert_1 = \lVert a\rVert_A + \lVert b\rVert_B$
$\lVert(a,b)\rVert_\infty = \max(\lVert a\rVert_A, \lVert b\rVert_B)$
I shall not prove that these norms induce the product topology, it is an easy exercise.
If we use these norms, we can readily verify continuity of additon and scalar multiplication. But it seems to me that this is by no means simpler than the above approach.