The converse is not true in general. For instance, if $X$ is compact, then any map $f:X\to Y$ for any $Y$ vacuously satisfies the second condition, since no net in $X$ goes to infinity. But not every map $f:X\to Y$ is proper. For instance, if $Y$ is $X$ with the indiscrete topology and $f$ is the identity map, then every subset of $Y$ is compact so this would be saying every subset of $X$ is compact which is certainly not true in general.
It is true if you require $Y$ to be Hausdorff. In that case, suppose $f:X\to Y$ is not proper; let $K\subseteq Y$ be compact such that $f^{-1}(K)$ is not compact. We can then find some net $(x_\alpha)$ in $f^{-1}(K)$ with no accumulation point in $f^{-1}(K)$. Since $Y$ is Hausdorff, $K$ is closed, so $f^{-1}(K)$ is closed, and so $(x_\alpha)$ actually has no accumulation point in $X$. It follows that $(x_\alpha)$ must go to infinity. But $f(x_\alpha)\in K$ for all $\alpha$, so $(f(x_\alpha))$ does not go to infinity.
As the comments got too long, I will make them in an answer. Let $P$ a nonconstant polynomial in the plane. Then:
1: $P$ is proper as a map $P:\mathbb{C} \to \mathbb{C}$
2: If we restrict $P$ to a domain $D$ and $\Omega=P(D)$ is the image domain ($P$ is an open map being analytic nonconstant so $\Omega$ is open, while it is connected by continuity), then $P$ is proper as a map $P: D \to \Omega$ if and only if $P(\partial D) \cap \Omega =\emptyset$
Proof: $|P(z)| \to \infty$ as $|z| \to \infty$ hence the preimage of bounded sets by $P$ is bounded. But in the plane, $K$ compact iff $K$ is bounded and closed and obviously $P^{-1}(K)$ is then bounded by the above argument and closed by continuity, hence it is compact, hence $P$ is proper as a plane map.
For 2, we notice that if $K \subset \Omega$, $K$ compact, then $P^{-1}(K) \cap D=L$ is bounded since $P^{-1}(K)$ is so, but $L$ is only relatively closed in $D$. However if $P(\partial D) \cap \Omega =\emptyset$ and $x_n \in L, x_n \to x$, $P(x_n) \to P(x) \in K$ so $x \in \bar D, P(x) \in \Omega$. As $\bar D = D \cup \partial D$ and we cannot have $x \in \partial D$ by hypothesis, it follows $x \in D$, hence $x \in L$, hence $L$ closed, so compact.
Conversely, if $P$ restricted to $D$ is proper and $x \in \partial D, P(x)=y \in \Omega$, then $P(z)=(z-x)^nQ(z)+y, n \ge 1, Q(x) \ne 0$ hence the preimage of a small closed disc centered at $y$ and included in $\Omega$ has a component that is a a compact neighborhood of $x$ in the plane, hence its intersection with $D$ is not compact which contradicts the fact that $P$ is proper restricted to $D$.
A simple example to show that this happens is $P(z)=z(z-1)$ on the open unit disc. Clearly $0 \in P(D)$ but a small closed disc centered at zero has as preimage in the open disc, two components, one around zero compact, but one that is the intersection of a compact neighborhood of $1$ with the open unit disc, hence non-compact.
In the examples of the OP, $z \to z^n$ on the discs centered at the origin, obviously the boundary condition is satisfied, so the maps are proper when restricted too, but if we shift the domain to a disc containing $1$ but having another root of unit of order $n$ on the boundary, the restricted maps are not proper any more
Best Answer
No, there’s no contradiction. It says that the identity is not a proper map of $U$ into $\mathbb C$, and that’s also what the criterion says: As you write, the sequence $p_i=1-\frac1i$ escapes to infinity in $U$, but $f(p_i)=1-\frac1i$ doesn’t escape to infinity in $\mathbb C$, so according to the criterion the map is indeed not proper.