Prove inequality holds for Single variable differentiable function

Question

I saw this question in another post and I proved it differently than the others who answered. I was wondering if my proof works.

Problem

Let $f: \mathbb{R} \to \mathbb{R}$ be a function. Suppose that $f$ is differentiable, that $f(0)=1$, and that $|f'(x)| \leq 1$ for all $x \in \mathbb{R}$. Prove that $|f(x)| \leq |x|+1$ for all $x \in \mathbb{R}$.

My Proof

Assume the conclusion didn't follow from the hypotheses, namely that $\exists x\in \mathbb{R}$ such that $|f(x)|>|x|+1$. We can show this for $x=0$. Then we get $|f(0)|>1 \Longrightarrow 1>1$, contradiction.

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And here's where I found the question Practice problem from Mean Value Theorem in Real Analysis

Answer 1

How can you say that $x=0$ is the point where $|f(0)|>|0|+1$?

We assume that there exists a point, $x_0$ such that $|f(x_0)|>|x_0|+1$, then we look at $$\frac{f(x_0)-f(0)}{x_0-0}$$ and you use this to prove that there is a point where $|f'(x)|>1$.

You can't just say "$x_0=0$", because the negative of all is exists, so it may be $0$, but it is also may be $10,5.4,777,\pi$ and so on.