We set $$\mathbb{R}[x] := \{ \alpha = a_nx^n + \cdots + a_1x + a_0 : \ a_i \in \mathbb{R} \ \ \ \forall i = 1,2\ldots , n\}.$$ for some non-negative integer $n$. It is well-known that $\mathbb{R}[x]$ becomes a vector space with addition and multiplication defined as follows.
(i) If $\alpha, \beta \in \mathbb{R}[x]$ with $\alpha = a_nx^n + \cdots + a_1x + a_0$ and $\beta = b_nx^n + \cdots + b_1x + b_0$, then $$\alpha + \beta = (a_n+b_n)x^n + \cdots + (a_1+b_1)x + (a_0+ b_0) \in \mathbb{R}[x] .$$ (ii) If $\alpha = a_nx^n + \cdots + a_1x + a_0 \in \mathbb{R}[x]$ and $c \in \mathbb{R}$, then $$c \alpha = a_ncx^n + \cdots + a_1cx + a_0c \in \mathbb{R}[x].$$
For each $\alpha = a_nx^n + \cdots + a_1x + a_0 \in \mathbb{R}[x]$, we define $$deg(\alpha) = \begin{cases} n & \text{ if } a_n \neq 0 \\ -\infty & \text{ if } \alpha = 0 \end{cases}.$$ Fix $q >1$. Define the map $||\cdot||_q:\mathbb{R}[x] \to \mathbb{R}_{\geq 0}$ by $$||\alpha||_q = q^{deg(\alpha)}.$$ The question is if $||\cdot||_q$ is a norm in $\mathbb{R}[x]$?
(a) $||\alpha||_q \geq 0$.
Let $\alpha \in \mathbb{R}[x],$ so $\alpha = a_nx^n + \cdots + a_1x + a_0$. Without loss of generality we can assume that $a_n \neq 0$. We then have: $$||\alpha||_q = q^{deg(\alpha)} = q^n,$$ but $q > 1$, so $q^n > 0$ and thus $||\alpha||_q \geq 0$. In particular, if the leading coefficient is zero, then $q^{deg(\alpha)} > 0$, because $q > 1$. Hence (a) is satisfied.
(b) $||\alpha||_q = 0 \Leftrightarrow \alpha = 0$.
"$\Rightarrow$"
Suppose $||\alpha||_q = 0$, we then have $q^{deg(\alpha)} = 0$. Since $q > 1$, the only possible way that $q^{deg(\alpha)} = 0$ is that $$\lim_{n \to \infty} \left(\frac{1}{q}\right)^n = 0,$$ this implies that $q^{-\infty} = 0$. Then $deg(\alpha) = -\infty$ which gives $\alpha = 0$.
"$\Leftarrow$"
Assume that $\alpha = 0$, then $deg(\alpha) = -\infty$ and $$||\alpha||_q = q^{deg(\alpha)} = q^{-\infty} = \lim_{n \to \infty} \left(\frac{1}{q}\right)^n = 0,$$ hence (b) is also satisfied.
(c) $||\alpha + \beta||_q \leq ||\alpha||_q + ||\beta||_q$.
Without loss of generality, we suppose that $$\alpha = a_nx^n + \cdots + a_1x + a_0$$ and $$\beta = b_mx^m + \cdots + b_1x + b_0,$$ where $m \leq n$. If $m = n$, then $\beta = b_nx^n + \cdots + b_1x + b_0$ and $$\alpha + \beta = (a_n+b_n)x^n + \cdots + (a_1+b_1)x + (a_0+ b_0),$$ and $$||\alpha + \beta||_q = q^{\deg(\alpha + \beta)} = q^{\deg(\alpha)} = q^{\deg( \beta)}.$$ Otherwise, set $m < n$. We then have $\beta = b_mx^m + \cdots + b_1x + b_0$, and $$\alpha + \beta = a_nx^n + \cdots + (a_m + b_m)x^m + \cdots + (a_1+b_1)x + (a_0+ b_0).$$ We thus have $$||\alpha + \beta||_q = q^{\deg(\alpha + \beta)} = q^{\deg(\alpha)}.$$ In a more general case, we have the inequality $$||\alpha + \beta||_q \leq \max\{||\alpha||_q, ||\beta||_q\}.$$ But does the triangle inequality holds in this case? We are given $q >1$ and in particular $q^k > 1$ for any $k \geq 1$. If $m = n$, we get $$||\alpha + \beta||_q = q^{\deg(\alpha + \beta)} = q^{\deg(\alpha)} < q^{\deg(\alpha)} + q^{\deg(\beta)} = ||\alpha||_q + ||\beta||_q.$$ Otherwise, we have $$||\alpha + \beta||_q \leq \max\{||\alpha||_q, ||\beta||_q\}.$$ We consider two cases, if $\max\{||\alpha||_q, ||\beta||_q\} = ||\alpha||_q$, then $$\max\{||\alpha||_q, ||\beta||_q\} \leq ||\alpha||_q + ||\beta||_q.$$ If $\max\{||\alpha||_q, ||\beta||_q\} = ||\beta||_q$, then $$\max\{||\alpha||_q, ||\beta||_q\} \leq ||\beta||_q + ||\alpha||_q.$$ Does this implies that the inequality $$||\alpha + \beta||_q \leq ||\alpha||_q + ||\beta||_q,$$ holds?
(d) Finally, for $\alpha \in \mathbb{R}[x]$ and $c \in \mathbb{R}$ ,we set $$c \alpha = a_ncx^n + \cdots + a_1cx + a_0c \in \mathbb{R}[x].$$ Now, $$||c \alpha||_q = q^{deg(c\alpha)} = q^{deg(\alpha)},$$ because $c$ is a scalar. Therefore $$||c \alpha||_q = ||\alpha||_q \neq |c| ||\alpha||_q,$$ unless $c = 1$. Thus the Positive Homogeneity property is not satisfied. Hence $(\mathbb{R}[x], ||\cdot||_q)$ is not a normed linear space.
Thanks in advance
It is correct,but in d), what happens if $c=0$?. The most delicate part was c), but you do it very well.
Addendum: there is a more general theory about this norm (maybe you already know). See a little piece $p$-adic numbers.