First it is shown that $f \left({x}\right)$ is bounded in the closed real interval $\left[{a \,.\,.\, b}\right]$.  {{AimForCont}} $f \left({x}\right)$ has no upper bound. Then for all $x \in \left[{a \,.\,.\, b}\right]$: :$\forall N \in \R: \exists x_n \in \left[{a \,.\,.\, b}\right]: f \left({\left \langle {x_n} \right \rangle}\right) > N$ We have that: :$\N \subset \R$ {{WLOG}}, we can consider $N \in \N$. Consider the sequence: :$\left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle_{n \in \N}$ and the sequence: :$\left \langle {N} \right \rangle_{N \mathop \in \N}$ Because: :$f \left({\left \langle {x_n} \right \rangle}\right) > N$ and: :$\left \langle {N} \right \rangle_{N \mathop \in \N} \to \infty$ it follows that: :$\left\langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle_{n \mathop \in \N} > \infty \implies \left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle_{n \in \N} \to \infty$ {{explain|$> \infty$?}} Consider the sequence: :$\left \langle {x_n} \right \rangle \in \left[{a \,.\,.\, b}\right]$ It is bounded: bounded above by $b$ and bounded below by $a$. So by the Bolzano-Weierstrass Theorem there exists a subsequence of $\left \langle {x_n} \right \rangle $: :$\left \langle {g_n} \right \rangle \to d$ Since $\left[{a \,.\,.\, b}\right]$ is closed: :$d \in \left[{a \,.\,.\, b}\right]$ By Continuity of Mapping between Metric Spaces by Convergent Sequence, then: :$f \left({x}\right)$ is continuous in $d$ and: :$f \left({\left \langle {g_n} \right \rangle}\right) = f \left({d}\right)$ But our first conlusion indicates that: :$\left \langle {f \left({x_n}\right)} \right \rangle_{n \in \N} \to \infty$ which contradicts: :$\left \langle {f \left({g_n}\right)} \right \rangle_{n \in \N} \to d$ A similar argument can be used to prove the lower bound. {{qed|lemma}} It remains to be proved that: :$\exists d \in \left[{a \,.\,.\, b}\right]: f \left({x}\right) \le f \left({d}\right)$ where $f \left({d}\right) = N$ (the maximum). It will be shown that: :$\forall n \in \R_{\ne 0}: N - 1/n < f \left({\left \langle {x_n} \right \rangle }\right) \le N$ as follows; {{WLOG}}, consider $n \in \N$. Let $I$ denote the codomain of $f \left[{a \,.\,.\, b}\right]$. Because $N$ is its maximum and $N - 1/n < N$: :$\forall n \in \N: \exists y_n \in \left[{a \,.\,.\, b}\right]: N - 1/n < y_n < N$ But $y_n \in I$, so: :$\forall n \in \N: \exists x_n \in \left[{a \,.\,.\, b}\right]$ That means: :$f \left({\left \langle {x_n} \right \rangle}\right) = y_n \implies  N - 1/n < f \left({x_n}\right) \le \ N$ {{qed|lemma}} It follows that: :$\left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle \to N$ Considering: :$\left \langle {N - 1/n} \right \rangle \to \ N$ as $n \to \infty$ and: :$\forall n \in \N: \left \langle {N} \right \rangle \to N$ by: : $\left \langle {N - 1/n} \right \rangle < \left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle \le N \implies \left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle \to \left \langle {N} \right \rangle$ {{explain|The grammatical sense of the above is unclear.}} Consider $\left \langle {x_n} \right \rangle$. Because it is bounded, by Bolzano-Weierstrass Theorem there exists a subsequence: :$\left \langle {s_n} \right \rangle$ that converges. Let it converge to $l$. Because $\left \langle {s_n} \right \rangle \in \left[{a \,.\,.\, b}\right]$ it follows that: :$l \in \left[{a \,.\,.\, b}\right]$. Finally, $f \left({x}\right) $ is continuous in $\left[{a \,.\,.\, b}\right]$. So, by Continuity of Mapping between Metric Spaces by Convergent Sequence: :$\left \langle {s_n} \right \rangle \to d \implies f \left({\left \langle {s_n} \right \rangle}\right) \to f \left({d}\right)$ But: :$\left \langle {f \left({x_n}\right)} \right \rangle \to N \implies \left \langle {f \left({s_n}\right)} \right \rangle_{n \in \N} \to N \iff f \left({d}\right) = N$ {{explain|Clarify, in that last compound proposition, what is being proved and how that conclusion is reached.}} {{qed}}