Regarding a holomorphic map from the unit disc to an convex domain

I am trying to understand the proof of the following result: Let $$M$$ be a convex domain in $$\mathbb{C}^n$$. Let $$\mathbb{D}$$ be the open unit disc in $$\mathbb{C}$$. If $$\phi: \mathbb{D}\longrightarrow \bar{M}$$ is holomorphic then either $$\phi( \mathbb{D})\subset M$$ or $$\phi( \mathbb{D})\subset \partial M$$.

The proof goes like this: Assume that $$z\in \phi( \mathbb{D})\cap \partial M$$. By convexity, we can choose a $$\mathbb{C}$$ linear functional $$l$$ such that Re $$l(z)>$$Re $$l(w)$$ for every $$w\in M$$. Hence $$l(z)\in(l\circ \phi)( \mathbb{D})$$ lies in the boundary of $$l\circ \phi( \mathbb{D})$$. By the open mapping theorem, $$l\circ \phi$$ is constant, So that $$\phi( \mathbb{D})$$ cannot contain points in $$M$$. Hence $$\phi( \mathbb{D})\subset \partial M$$.

I was able to understand the proof till the part where we get $$l\circ \phi$$ is constant. But why does it imply that $$\phi( \mathbb{D})$$ cannot contain points in $$M$$?

If there exists $$w\in \mathbb{D}$$ such that $$\phi(w)\in M,$$ then $$\mathrm{Re}\; l\circ \phi(w)<\mathrm{Re}\;l(z)$$ by assumption. However, $$z\in \phi(\mathbb{D})$$ and thus, $$l\circ \phi$$ cannot be constant.