Show the space of $(\alpha,\mu)$-Hölder continuous functions is not closed under scalar multiplication. CONTEXT: I want to show $F_{\alpha,\mu}$ is not a linear subspace of $C([0,1],\mathbb{R})$.
We call $f:[0,1]\to\mathbb{R}$ an "$(\alpha,\mu)$-Hölder continuous function" if 
$$
\exists_{\alpha,\mu>0} \forall_{x,y\in[0,1]}: |f(x)-f(y)|\leq\mu\cdot|x-y|^{\alpha}.
$$
Let $F_{\alpha,\mu}$ be the set of $(\alpha,\mu)$-Hölder continuous functions.
I have previously shown that if $f\in F_{\alpha,\mu}$, then $p f\in F_{\alpha,|p|\mu}$, for all $p\in\mathbb{R}$. (1)
I now want to show that $F_{\alpha,\mu}$ is not closed under scalar multiplication. To do this, I assume it is closed under scalar multiplication. Proof by contradiction.
Suppose $F_{\alpha,\mu}$ is closed under scalar multiplication for all $\alpha,\mu$. Let $f\in F_{\alpha,\mu}$. Then also $pf\in F_{\alpha,\mu}$ (because of closure). In the same way $\frac{1}{p}pf\in F_{\alpha,\mu}$, but because $pf\in F_{\alpha,\mu}$ we have by our previous result (1) that $f=\frac{1}{p}pf \in F_{\alpha,|\frac{1}{p}|\mu}$.
So we conclude that when $F_{\alpha,\mu}$ is closed under scalar multiplication, we have that for all $f\in F{\alpha,\mu}$ that also $f\in F{\alpha,p\mu}$, for all $p\in\mathbb{R}$.
Returning to the definition I think this means that we now have
$$
\exists_{\alpha>0} \forall_{\mu\in\mathbb{R}\backslash\{0\}}\forall_{x,y\in[0,1]}:|f(x)-f(y)|\leq \mu\cdot|x-y|^{\alpha}.
$$
Because this is true for all $\mu$, also for $\mu$ approaching to zero, we have that $|f(x)-f(y)|=0$ and thus that $f$ is constant. This is the line I cannot prove rigorously. I think it is true because $\alpha$ is fixed.
We conclude that all Hölder continuous functions must be constant. But $f(x)=x$ is (1,1) Hölder continuous. Therefore $F_{\alpha,\mu}$ cannot be closed under scalar multiplication, this was my false assumption. So $F_{\alpha,\mu}$ is not a linear subspace of $C([0,1],\mathbb{R})$.
Is this line of reasoning correct? And how do I fix the vagueness that I put in bold?
 A: The point is that if $t$ is a real number such that $|t| \le \mu$ for all $\mu > 0$ then $t = 0$.  Well, this is rather obvious: if $t \ne 0$ you could take $\mu = |t|/2$...
A: Your argument works to show that it is not true that $F_{\alpha,\mu}$ is closed under scalar multiplication for all $\alpha$ and $\mu$.  For the step you asked about, there really isn't more you need to say: $|x-y|^\alpha$ is just some number $C$ and you know $|f(x)-f(y)|\leq \mu C$ for all $\mu> 0$, which implies $|f(x)-f(y)|=0$.  To be very explicit, if $|f(x)-f(y)|$ were positive, you could take $\mu=\frac{|f(x)-f(y)|}{2C}$ to get a contradiction (assuming $C\neq 0$, since the $C=0$ case is trivial).
Note though that this does not prove what you wanted to prove, which is that $F_{\alpha,\mu}$ is not closed under scalar multiplication for all $\alpha$ and $\mu$.  All your argument shows is that there is some $\alpha$ and $\mu$ for which it is not closed under scalar multiplication.  To be more precise, what your argument really shows is that if $F_{\alpha,\mu}$ is closed under scalar multiplication for some particular $\alpha$ and $\mu$ then $F_{\alpha,\mu}$ consists only of constant functions.  So, your example $f(x)=x$ being in $F_{1,1}$ only shows that $F_{1,1}$ is not closed under scalar multiplication.  To show it in general, you need to consider arbitrary $\alpha$ and $\mu$ and find a nonconstant function in $F_{\alpha,\mu}$.
