calculus - Proof that a function is bounded

The question :

Let $f:[1,\infty)\to \mathbb{R}$
be a continuous function such that $\underset{x\rightarrow\infty}{\lim}f(x)=L$

Prove that the function is bounded.

My try :

By definition a continuous function $f:[1,\infty)\rightarrow\mathbb{R}$
1. $f$
is continuous in $(1,\infty)$

i.e $\forall x_{0}>1\underset{x\rightarrow x_{0}}{\lim}f(x)=f(x_{0})$

2. $f$
   continuous at $1^{+}$
   i.e $\underset{x\rightarrow1^{+}}{\lim}f(x)=f(1)$

as stated in the question's contitions :$\underset{x\rightarrow\infty}{\lim}f(x)=L\iff\forall\varepsilon>0\,\exists M\in\mathbb{R}: x>M\rightarrow\left|f(x)-L\right|<\varepsilon$

Set $\varepsilon=\left|f(1)-L\right|$

Then exists $M$
such that $x>M\rightarrow\left|f(x)-L\right|\leq\left|f(1)-L\right|$
including when $x=1$

Set $\left|f(1)-L\right|=K$

This also implies that$M\geq1$

Thus we get $x\geq1\rightarrow\left|f(x)-L\right|\leq K \iff x\geq1\rightarrow L-K\leq f(x)\leq L+K$

Thus $f$
is bounded.

However I feel that this proof doesnt work and I do not fully understand what I have done here actually (just tried to replicate the lecture notes of my professor)

I want to understand this question and the correct way to answer it.
Assistance will be greatly appreciated.

Answer

Since the comments seem to indicate that the OP is not familiar with the theorem that a continuous function $f$ is bounded on a compact set, here is a simple proof for the special case of a closed bounded interval $[a,b] \subset \mathbb R$.

Suppose for a contradiction that $f$ is not bounded on $[a,b]$. Therefore it must be either unbounded above or unbounded below. Without loss of generality, assume $f$ is unbounded above. (Otherwise replace $f$ with $-f$.)

Now divide the interval into two subintervals, $[a + (a+b)/2]$ and $[(a+b)/2, b]$. Now $f$ must be unbounded on one of these subintervals. Repeating this procedure, we identify a sequence of closed bounded intervals $[a,b] = I_0 \supset I_1 \supset I_2 \supset \cdots$ such that the length of $I_n$ is $(b-a)/2^n$, and $f$ is unbounded on each of these intervals.

This means that we can choose points $x_0 \in I_0$, $x_1 \in I_1$, $x_2 \in I_2$, etc. such that $f(x_n) > n$ for every $n$.

Now each $I_n$ contains every $x_k$ for $k \geq n$, and from this we can easily conclude that $x_k$ converges to some limit $x$. Since each $x_n$ is in $[a,b]$ and $[a,b]$ is closed, it contains all of its limit points, hence $x\in [a,b]$.

By continuity of $f$, we must have
$$\lim_{n \to \infty}f(x_n) = f(x)$$
but this is impossible since $f(x_n) > n$ for every $n$.

Our assumption that $f$ is unbounded on $[a,b]$ is untenable, so $f$ must be bounded after all.