Here's some intuition for the problem:
First, think about working coordinate-wise, one $n$ at a time. What does it mean for $(1-t)x_n + ty_n$ to be continuous (at $t_0$)?
It means for all $\epsilon > 0$, there must be a $\delta >0$ such that if $0<|t-t_0|<\delta$, then $|f(t_0) - f(t)|<\epsilon$.
In fact, we can choose $\delta = \epsilon/|x_n - y_n|$ when $x_n\neq y_n$. To see this, just compute: \begin{align*} |f(t_0) - f(t)| &= |(1-t_0)x_n + t_0y_n - (1-t)x_n - ty_n| \\\ &= |(t-t_0)(x_n - y_n)| \\\ &= |t_0-t||x_n - y_n|\\\ &< \frac{\epsilon}{|x_n-y_n|}|x_n-y_n|\\\ &= \epsilon \end{align*}
So we can choose this $\delta$, but we learn something more: this is the best possible $\delta$. For if $|t-t_0| = \delta$, then the same calculation shows that $|f(t) - f(t_0)| = \epsilon \not<\epsilon$.
What happens if $x_n = y_n$? In this case, $(1-t)x_n + y_n = x_n$, so it's constant. This means that given any $\epsilon$, we may choose any $\delta$ we want.
Now, in the box topology, we get to pick a different $\epsilon_n$ on each factor, but the $\delta$ that we pick must work for all $\epsilon_n$ simultaneously. This leads us to the idea of the solution: If there are only finitely many $x_n\neq y_n$, then no matter what $\epsilon_n$s we choose, we can choose the $\delta_n$s as above, and let $\delta = \min\{\delta_n\}$.
On the other hand, if we have $x_n\neq y_n$ infinitely often, then, given $\epsilon_n$s the same approach as in the previous paragraph leads us to try $\delta = \inf\{\delta_n\}$. In fact, if anything works, it must be this $\delta$, owing to the fact that each $\delta_n$ above is the best possible. Note that we're taking an $\inf$ instead of a $\max$ because there are infinitely many $\delta_n$s. Now, if this $\inf$ is $0$, then the function $f$ fails to be continuous.
So, in order to force $\delta = 0$, we need the $\delta_n$s to decrease to $0$, at least when considering the $n$ where $x_n\neq y_n$. To do this, pick $\epsilon_n = \frac{1}{n} |x_n-y_n|$ (if $x_n = y_n$, just pick $\epsilon = 1$).
Then our above formula for $\delta_n$ gives $\delta_n = \frac{\epsilon}{|x_n-y_n|} = \frac{1}{n}$. Then, $\delta = \inf\{\delta_n\} = 0$.