Definition of the norm of a bounded linear operator. I have a somewhat basic but confusing question regarding the definition of the norm for bounded linear operator.
Suppose $f$ is a bounded linear operator, that is, there exists $M>0$ such that $\|f(x)\| \leq M\|x\|$. We define $\mathcal{M}=\{M\geq 0:\|f(x)\|\leq M\|x\|, \forall x\}$
$$\|f\|:=\inf\{M\geq 0:\|f(x)\|\leq M\|x\|, \forall x\}=\inf\{\mathcal{M}\}.$$
Now, the question is: how do we know that $\|f\|\in \mathcal{M}$, i.e. that $\|f(x)\|\leq \|f\|\|x\|$ for all $x$?
This might seem a rather dumb question, but I've noticed a lot of books skirt around the technicalities of this, and even when they don't, the arguments don't strike me as convincing. The argument I've seen a lot goes like this:
Since $\|f\| =\inf\{\mathcal{M}\}$, for all $n\in\mathbb{N}$, $$\|f\|+\frac1n \in \mathcal{M}.$$ Otherwise $\|f\|$ would not be the infimum. Therefore
$$\|f(x)\| \leq \left( \|f\|+\frac1n \right)\|x\|,$$
for all $x$ and for all $n\in \mathbb{N}$. Letting $n\rightarrow \infty$ while keeping everything else fixed we obtain:
$$\|f(x)\| \leq \|f\|\|x\|.$$ 
How exactly is $\|f(x)\| \leq \|f\|\|x\|$ a consequence of $\|f(x)\| \leq \left( \|f\|+\frac1n \right)\|x\|$? 
I tried obtaining a proof by contradiction: let us replace $\frac1n$ by $\epsilon>0$ and let's assume that the latter is true but the conclusion is not. This would imply the existence of $x_0$ such that 
$$\|f(x_0)\| > \|f\|\|x_0\|$$ while simultaneusly having
$$\left( \|f\|+\epsilon \right)\|x_0\|\geq \|f(x_0)\|.$$
Without loss of generality we can assume $x_0\neq 0$, and putting this all together yields:
$$\|f\|+\epsilon> \|f\|,$$
which is not a contradiction.
More generally, the fact that $\|f\|+\epsilon \in \mathcal{M}$ amounts to saying that $\|f\|$ is a limit (or accumulation, depending on your definition) point of $\mathcal{M}$. It does not follow that the limit point must belong to the set in question (that would be saying $\mathcal{M}$ is closed (which is what we're trying to prove).
Thanks in advance for any insights into this.
 A: Let's say you assumed the existence of $x_0$ such that
$$
\Vert f(x_0) \Vert > \Vert f\Vert \Vert x_0\Vert.
$$
Then you know there exists an $\epsilon \in\mathbb{R}_{>0}$ with
$$
\Vert f(x_0) \Vert > (\Vert f\Vert +\epsilon) \Vert x_0\Vert
$$
because the inequality holds strictly (so we can make the right side a little bigger and the inequality still holds). This is basically how you started.
By definition of $\Vert f \Vert$ we know that for all $n\in\mathbb{N}$
$$
\Vert f(x_0)\Vert \leq \left(\Vert f\Vert + \frac{1}{n}\right) \Vert x_0\Vert.
$$
Now we use that there exists an $n\in\mathbb{N}$ with $\frac{1}{n} < \epsilon$ (this is possible because there always exists $n\in\mathbb{N}$ with $n > \frac{1}{\epsilon}$). From that we get
$$
\Vert f(x_0)\Vert \leq \left(\Vert f\Vert + \frac{1}{n}\right) \Vert x_0\Vert < \left(\Vert f\Vert + \epsilon\right) \Vert x_0\Vert.
$$
But now we have
$$
(\Vert f\Vert +\epsilon) \Vert x_0\Vert < \Vert f(x_0) \Vert < \left(\Vert f\Vert + \epsilon\right) \Vert x_0\Vert,
$$
which is a contradiction.
