Let's extend your PDE with some initial and boundary conditions
$$
u_t = -u_x, x \in [0, 1]\\
u\big|_{x=0} = \beta(t)\\
u\big|_{t=0} = u_0(x)
$$
By discretizing the time, you've obtained the system of ODEs with initial conditions:
$$\begin{aligned}
&u_{n+1}(x) + h u'_{n+1}(x) = u_{n}(x)\\
&u_{n+1}(0) = \beta((n+1)h).
\end{aligned}
$$
From this point, there are two ways to show the absolute stability of the scheme.
Maximum principle. One can show that $||u_{n+1}||_C \leq \max(||\beta||_C, ||u_n||_C)$. Proof:
Provided that $u_{n+1}(x) \in C^1[0,1]$ the global maximum of $|u_{n+1}(x)|$ can be reached either at $x = 0$, or at $x = 1$ or at some point $0 < x < 1$ with $u'_{n+1}(x) = 0$. If the maximum is at $x = 0$ then $||u_{n+1}||_C \leq ||\beta||_C$. If the maximum is at $0 < x < 1$ then
$$
||u_{n+1}||_C = |u_{n+1}(x)| = |u_{n}(x)| \leq ||u_n||_C
$$
due to $u_{n+1}'(x) = 0$. Finally, if the maximum is at $x = 1$, then
$$
\operatorname{sign}(u_{n+1}(1)) u'_{n+1}(1) \geq 0
$$
so
$$
|u_{n+1}(1)| + h\operatorname{sign}(u_{n+1}(1)) u'_{n+1}(1) =
\operatorname{sign}(u_{n+1}(1))u_n(1).
$$
Since the left side is nonnegative, the right side should be $|u_n(1)|$. Again
$$
|u_{n+1}(1)| \leq |u_n(1)| \leq ||u_n||_C.
$$
So we have proved that
$$
||u_{n+1}||_C \leq \max(||u_n||_C, ||\beta||_C) \leq \max(||u_0||_C, ||\beta||_C)
$$
i.e. the scheme is stable under $||\cdot||_C$ norm.
Operator norm theory. First, let's simplify the problem a bit:
$$
u_{n+1}(x) + h u_{n+1}'(x) = u_n(x)\\
u_{n+1}(0) = \beta((n+1)h)
$$
by introducing $y_{n+1}(x) = u_{n+1}(x) - \beta((n+1)h)$:
$$
y_{n+1}(x) + h y_{n+1}'(x) = y_n(x) - \beta((n+1)h) + \beta(nh)\\
y_{n+1}(0) = 0
$$
Now every $y_{n+1}(x)$ belongs to subspace of $C^1[0,1]$ with $y_{n+1}(0) = 0$ condition.
Let's rewrite the operator scheme in form
$$
B y_{n+1} = y_n + \varphi_n
$$
with $A$ and $B$ being some operators.
$$
B = I + h D\\
\varphi_n = \beta(nh)-\beta((n+1)h).
$$
Let the inner product $(u,v)$ be the
$$
(u,v) = \int_{0}^1 u(x)v(x) dx.
$$
Then
$$
(By,y) = (y,y) + h(Dy,y)\\
(Dy,y) = \int_0^1 y y'dx = \int_{y(0)}^{y(1)} y dy = \frac{y^2(1)}{2} - \frac{y^2(0)}{2} = \frac{y^2(1)}{2}\\
(By,y) = (y,y) + h\frac{y^2(1)}{2} \equiv ||y||_B^2 \geq (y,y) \equiv ||y||^2\\
(Dy,y) = \frac{y^2(1)}{2}
$$
so $B$ is positive definite and $D$ is positive-semidefinite. Thus there exists $B^{-1}$. We would like to find some Hilbert space where $B^{-1}$ has norm not greater than 1.
Using the inequality $||y||^2 \leq ||y||^2_B$
$$
||B^{-1}x||^2 \leq ||B^{-1}x||^2_B = (BB^{-1}x,B^{-1}x) = (x, B^{-1}x) \leq ||x||\cdot||B^{-1}x||\\
||B^{-1}x|| \leq ||x||
$$
we can conclude that $||B^{-1}|| \leq 1$ (so the space has simply $(u,v)$ for its inner product).
Now
$$
y_{n+1} = B^{-1}(y_n + \varphi_n)\\
||y_{n+1}|| \leq ||B^{-1}||(||y_n||+||\varphi||) \leq ||y_n|| + ||\varphi_n||\\
||y_{n+1}|| \leq ||y_0|| + \sum_{k=0}^{n} ||\varphi_k||
$$
Since $\varphi_k(x) = \beta(nh)-\beta((n+1)h)$,
$$
||\varphi_k|| = |\beta((n+1)h)-\beta(nh)|\sqrt{\int_0^1 dx} = |\beta((n+1)h)-\beta(nh)| \leq h \max_{0 \leq t \leq (n+1)h} |\beta'(t)|
$$
$$
||y_{n+1}|| \leq ||y_0|| + \int_0^{(n+1)h} |\beta'(t)|dt
$$
Thus again we've proved that the scheme is stable, but in $L_2$ norm this time.
