There is no such map. For convenience, I will use the upper half-plane $\mathbb{H}$ instead of the unit disk.
Suppose $f\colon \mathbb{H}\to\mathbb{C}$ is a proper holomorphic map.
Note first that, for each sequence $\{z_n\}$ in $\mathbb{H}$ that converges to a point $p$ on the real axis, the image sequence $\{f(z_n)\}$ must must converge to $\infty$ on the Riemann sphere. To prove this, observe that if $\overline{D}_R$ is the closed disk of radius $R$ in $\mathbb{C}$ centered at the origin, then $\overline{D}_R$ is compact, and hence $f^{-1}(\overline{D}_R)$ is a compact subset of $\mathbb{H}$. Then $\mathbb{C}-f^{-1}(\overline{D}_R)$ is an open neighborhood of $p$, so $z_n$ lies in the complement of $f^{-1}(\overline{D}_R)$ for all but finitely many $n$. This holds for all $R>0$, which proves the claim.
Now, since $f$ is a nonzero holomorphic function, the zeroes of $f$ form a discrete subset of $\mathbb{H}$. Since $f$ is proper, the set of zeroes must be compact, and therefore there are only finitely many zeroes. Let $D$ be any open disk centered on the real axis that does not contain any zeroes of $f$, and consider the holomorphic function $g(z) = 1/f(z)$ on $D\cap\mathbb{H}$.
Now, $g$ has the property that $g(z_n) \to 0$ for any sequence $z_n \in D\cap\mathbb{H}$ converging to a point on the real axis. Thus $g$ extends continuously to $D\cap\overline{\mathbb{H}}$ by setting $g(x) = 0$ for $x \in D\cap\mathbb{R}$. By the Schwarz reflection principle, $g$ now extends to a holomorphic function on all of $D$, which is nonsense since $g$ is zero on $D\cap\mathbb{R}$.
Best Answer
Such a map need not be proper. A simple example is given by the map $f \colon z \mapsto z^2$ with $Y = \mathbb{C}\setminus \{0\}$ and $X$ any connected open subset of $Y$ such that $f(X) = Y$. Standard choices for $X$ are $\mathbb{C} \setminus (-\infty,0]$ or a three-quarter plane. A maximal $X$ is for example $Y \setminus \{1\}$.
Then if $z_0$ is a boundary point of $X$ in $Y$, the preimage of a compact neighbourhood of $f(z_0)$ isn't compact.