We want to prove that function $f$, which is continuous on $[a,b]$, is bounded from above. There is one step in the proof that I don't understand.
If $f$ is not bounded from above, then for every positive integer $n$ we can find a number $x_n$ in the interval such that $f(x_n) \gt n$. The sequence of such $x_n$ has a point of accumulation $C$ in the interval by the Weierstrass-Bolzano theorem. By continuity, given $1$, there exists $\delta$ such that if $x \in [a,b]$ and $|x - C| \lt \delta$, then $|f(x) - f(C)| \lt 1$. In particular, $|f(x_n)| - |f(C)| \le |f(x_n) - f(C)| \le 1$, whence $n \lt f(x_n) \le 1 + |f(C)|$. This is a contradiction for $n$ sufficiently large, thus proving that $f$ is bound from above.
I don't see any contradiction. What's the problem with $f(x_n)$ being not larger than $1 + |f(C)|$ for every $n$?
We get a contradiction when $n>1+|f(C)|$.