Intermediate Value Theorem: If $f:[a,b]_{\mathcal{U}_{[a,b]}} \rightarrow \mathbb{R}_{\mathcal{U}}$ is continuous and if c is any real number between $f(a)$ and $f(b)$, then there exists a number $z \in (a,b)$ such that $f(z) = c$.
Remark: $\mathcal{U}$ indicates the usual topology on $\mathbb{R}$ and $\mathcal{U}_{[a,b]}$ is the subspace topology on A.
Proof: Suppose that $\forall x \in (a,b)$, $f(x) \neq c$. Since $f$ is continuous, then $Im(f)$ is connected. Consider the following disjoint sets: $$A = (-\infty,c) \cap Im(f) \qquad and \qquad (c,+\infty) \cap Im(f)$$ Since $(-\infty,c)$ and $(c,+\infty)$ are both $\mathcal{U}$-open, therefore $A$ and $B$ are $\mathcal{U}_{Im(f)}$-open. Observe that $f(x) \neq c$ for all $x \in [a,b]$, i.e., $c \notin Im(f)$. It follows that $A \amalg B = Im(f)$. So by definition $Im(f)$ is disconnected, contradiction.
Is this a valid proof?