Here is a sketch of how the proof might go.

If $f:[0,1]\rightarrow\mathbb R$ is continuous but not bounded then the sets $S_n=[0,1]\setminus f^{-1}[(-n,n)]$ are closed with $S_n\ne\emptyset$ for all $n\in\mathbb N$. According to the definition of continuous function the sets $f^{-1}[(-n,n)]$ are represented as $\Sigma_1$ sets of (endpoints of) rational intervals ($\Sigma_1$ definable in the model). So the closed dyadic rational intervals contained in $S_n$ for $n\in\mathbb N$ can be represented as an infinite $\Pi_1$ (or by a standard trick, equivalently $\Delta_1$) tree, which by Weak König's Lemma must have a path, so $\cap_n S_n\ne\emptyset$ which is absurd.

Here is a sketch of how the proof might go.

If $f:[0,1]\rightarrow\mathbb R$ is continuous but not bounded then the sets $S_n=[0,1]\setminus f^{-1}[(-n,n)]$ are closed with $S_n\ne\emptyset$ for all $n\in\mathbb N$. According to the definition of continuous function the sets $f^{-1}[(-n,n)]$ are represented as $\Sigma_1$ sets of (endpoints of) rational intervals ($\Sigma_1$ definable in the model). So the closed dyadic rational intervals intersecting $S_n$ for $n\in\mathbb N$ can be represented as an infinite $\Pi_1$ (or by a standard trick, equivalently $\Delta_1$) tree, which by Weak König's Lemma must have a path, so $\cap_n S_n\ne\emptyset$ which is absurd.

Here is a sketch of how the proof might go.

If $f:[0,1]\rightarrow\mathbb R$ is continuous but not bounded then the sets $S_n=[0,1]\setminus f^{-1}[(-n,n)]$ are closed with $\cap_n S_n=\emptyset$, but $\cap_{n\le N}S_n\ne\emptyset$ for all $N\in\mathbb N$. According to the definition of continuous function the sets $f^{-1}[(-n,n)]$ are represented as $\Sigma_1$ sets of (endpoints of) rational intervals ($\Sigma_1$ definable in the model). So the closed dyadic rational intervals contained in $\cap_{n\le N} S_n$ can be represented as an infinite $\Delta_1$ tree, which by Weak König's Lemma must have a path, so $\cap_n S_n\ne\emptyset$ after all.

Here is a sketch of how the proof might go.

If $f:[0,1]\rightarrow\mathbb R$ is continuous but not bounded then the sets $S_n=[0,1]\setminus f^{-1}[(-n,n)]$ are closed with $S_n\ne\emptyset$ for all $n\in\mathbb N$. According to the definition of continuous function the sets $f^{-1}[(-n,n)]$ are represented as $\Sigma_1$ sets of (endpoints of) rational intervals ($\Sigma_1$ definable in the model). So the closed dyadic rational intervals contained in $S_n$ for $n\in\mathbb N$ can be represented as an infinite $\Pi_1$ (or by a standard trick, equivalently $\Delta_1$) tree, which by Weak König's Lemma must have a path, so $\cap_n S_n\ne\emptyset$ which is absurd.