Show continuity of $f:\mathbb R \to \mathbb R^{2 \times 2}$ $f:\mathbb R \to \mathbb R^{2 \times 2}$
$f(x)=\begin{pmatrix}1&&0\\1-x&&x\end{pmatrix}$
Now, I know that $f$ is continuous because it is continuous entry wise, but I want to show it directly.
Let $|x-y|<\delta$
$\|f(x)-f(y)\|=\|\begin{pmatrix}0&&0\\y-x&&x-y\end{pmatrix}\|=\sup\{\|\begin{pmatrix}0&&0\\y-x&&x-y\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix}\|:v \in \mathbb R^2, \|v\|\le 1\}=\sup\{\|\begin{pmatrix}0\\(y-x)v_1+(x-y)v_2\end{pmatrix}\|:v \in \mathbb R^2, \|v\|\le 1\}$
How to continue?
 A: $||\begin{pmatrix}y-x\\x-y\end{pmatrix}\|= \sqrt{2}|x-y|$.
Can you proceed ?
A: By Cauchy-Schwarz, and the fact that $\|\binom{0}{a}\| = |a|$, $$\left\|\begin{pmatrix}0\\(y-x)v_1+(x-y)v_2\end{pmatrix}\right\| = \left|\binom{y-x}{x-y} \cdot v\right| \le   \left\|\binom{y-x}{x-y}\right\|\| v\| =\sqrt2|y-x|\|v\|  $$
Take $\sup_{v \in \mathbb R^2, \|v\|\le 1}$ to conclude.
A: Since all matrix norms are equivalent, you can consider the Frobenius norm on $\mathbb{R}^{2 \times 2}$ instead of the operator norm.
We have
$$\|f(x)-f(y)\|_F = \left\|\begin{pmatrix} 0 & 0 \\ y-x & x-y\end{pmatrix}\right\|_F = \sqrt{(y-x)^2 + (x-y)^2} = \sqrt2 |x-y|$$
so $f$ is continuous.
A: View $f$ as a map $f:\mathbb{R}\to \mathbb{R}^4$. 
Such a map is continuous iff all its components are continuous, which is clearly the case here.
If you insist on a direct proof, just redo the proof of that fact in this special case. 
A: For any given norm $\|\cdot\|$, 
$$
\|f(x)-f(y)\| = |x-y|\underbrace{\left\|\left(\begin{array}{cc}0&0\\-1&1\end{array}\right)\right\|}_{n}.
$$
For $\varepsilon>0$, $|x-y|<\delta = \varepsilon/(2n)$ implies $\|f(x)-f(y)\|< \varepsilon/2<\varepsilon$.
