Let $\gamma = (\gamma_1(t), \gamma_2(t))$ be any divergent curve. We need to show its length is infinite. To that end, let $N$ be any positive number. We'll show that the length of $\gamma$ is bigger than $N$.
Note that the length of $\gamma$ is $L = \int_0^\infty \frac{1}{\gamma_2(t)} \sqrt{\gamma'_1(t)^2 + \gamma'_2(t)^2} dt$
To that end, consider the rectangle $[\gamma(0) - L, \gamma(0) + L] \times [\epsilon, M]$ where $L$, $M$, and $\epsilon$ are to be determined. Note that this rectangle is compact since the topology is the usual one on the plane.
Since $\gamma(0)$ is in the rectangle, there is some maximum time $t_0$ such that $\gamma(t)$ is not in the rectangle for $t>t_0$. By considering a linear reparameterization $t' = \frac{t}{t_0}$ (which doesn't change lengths), we may assume wlog that $t_0 = 1$.
Now, there are 4 cases to consider. Either $\gamma(1)$ is on the top, bottom, left, or right edge of the rectangle. (If it hits a corner, say, the top left corner, then it's on top edge and left edge simultaneously).
Suppose $\gamma(1)$ hits the top or bottom. Then we have \begin{align*} \int_0^1 \frac{1}{\gamma_2(t)} \sqrt{\gamma_1'(t)^2 + \gamma_2'(t)^2} dt &\geq \int_0^1 \frac{|\gamma_2'(t)|}{\gamma_2(t)} dt\\ &\geq \left|\int_0^1 \frac{\gamma_2'(t)}{\gamma_2(t)} dt\right| \\ &= \left|\left.\ln(\gamma_2(t))\right]_0^1\right|\\ &= |\ln(\gamma_2(1)) - \ln(\gamma_2(0))|.\end{align*}
Now, if $\gamma(1)$ hits the top, then $\gamma_2(1) = M$, so the last line is $|\ln(M) - \ln(\gamma_1)|.$ Clearly, by picking $M$ large enough, we can make this bigger than $N$. If $\gamma(1)$ hits the bottom, then the last line is $|\ln \epsilon - \ln(\gamma(1))|$ and again, by choosing $\epsilon$ sufficiently close to $0$, we can make this bigger than $N$.
Finally, assume $\gamma(1)$ hits one of the sides of the rectangle. Then we have \begin{align*}\int_0^1 \frac{1}{\gamma_2(t)} \sqrt{\gamma_1'(t)^2 + \gamma_2'(t)^2} dt &\geq \int_0^1 \frac{|\gamma_1'(t)|}{\gamma_2(t)} dt\\ &\geq \int_0^1 \frac{|\gamma_1'(t)|}{M} \\ &\geq \left| \frac{1}{M}\int_0^1 \gamma_1'(t)dt \right| \\ &= \left| \left.\frac{1}{M} \gamma_1(t)\right]_0^1 \right| \\ &= \left| \frac{1}{M}(\gamma(0)\pm L - \gamma(0))\right| \\ &= \frac{L}{M}. \end{align*} So again, it's clear that if $L$ is big enough, then the length of $\gamma$ is at least $N$.