I find it cleanest to think in terms of Lipschitzness, rather than the $\varepsilon$-$\delta$ definition of continuity.
Let $(X, d)$ be a metric space.
For any $a \in X$, the map $d (a, \cdot) : X \to \mathbb R$ is Lipschitz, and hence is continuous.
The map $d : X \times X \to \mathbb R$ is Lipschitz w.r.t. the product metric, and hence is continuous.
I will prove (2.) and leave (1.) as a simpler exercise. (In fact, (2.) also directly implies (1.).) Fix $(x, y), (z, w) \in X \times X$. Then $ \begin{array}{rll} |d(x,y) - d(z,w)| &\leqslant |d(x,y) - d(z,y)| + |d(z,y) - d(z,w)| & \\ &\leqslant d(x, z) + d(y, w) & \text{(triangle inequality on } d \text{)} \\ &\leqslant 2 \max \{ d(x, z) , d(y, w) \} & \\ &= 2 d_{\infty} ((x, y) , (z, w)), & \end{array} $ showing that $d$ is Lipschitz. (Here we have assumed the “max” metric on the product; certainly other choices are possible but they turn out to be equivalent.)
Coming to the OP's question, to prove that a norm on a linear space is continuous w.r.t. itself, notice that the linear space is also a metric space with $d(x,y) = \| x - y \|$. In other words, $\| x \|$ is just the distance of $x$ from the origin; hence applying item (1.) above (with $a = 0$) shows the continuity of the norm.
A point to ponder: Equivalent norms. The above discussion might suggest that continuity of a norm is not an interesting thing to study. However this question can be modified a little to give rise to a quite fruitful concept:
Given two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ on a linear space $X$, are they continuous with respect to each other? I.e., when $\| \cdot \|_1$ is viewed as a function, is it continuous w.r.t. the norm $\| \cdot \|_2$ and vice versa?
It can be shown that this is true if and only if there exist numbers $0 < \alpha \leqslant \beta < \infty$ such that $ \alpha \| v \|_1 \leqslant \| v \|_2 \leqslant \beta \| v \|_1 $ for all vectors $v$. In this case, we say that the two norms are equivalent.