Solution:

Answer: $2 \sqrt{5}$

Let $l_{1}$ and $l_{2}$ be the lines as labeled in the above diagram. First, suppose Knot visits $l_{1}$ first, at point $P_{1}$, then $l_{2}$, at point $P_{2}$. Let $K^{\prime}$ be the reflection of $K$ over $l_{1}$, and let $T^{\prime}$ be the reflection of $T$ over $l_{2}$. The length of Knot's path is at least
$$K P_{1} + P_{1} P_{2} + P_{2} T = K^{\prime} P_{1} + P_{1} P_{2} + P_{2} T^{\prime} \geq K^{\prime} T^{\prime}$$
by the Triangle Inequality (This bound can be achieved by taking $P_{1}, P_{2}$ to be the intersections of $K^{\prime} T^{\prime}$ with $l_{1}, l_{2}$, respectively.) Also, note that $\measuredangle K^{\prime} O T^{\prime} = 90^{\circ}$, so that $K^{\prime} T^{\prime} = 2 \sqrt{5}$.

Now, suppose Knot instead visits $l_{2}$ first, at point $Q_{2}$, then $l_{1}$, at point $Q_{1}$. Letting $K^{\prime\prime}$ be the reflection of $K$ over $l_{2}$ and $T^{\prime\prime}$ be the reflection of $T$ over $l_{1}$, by similar logic to before the length of his path is at least the length of $K^{\prime\prime} T^{\prime\prime}$. However, by inspection $K^{\prime\prime} T^{\prime\prime} > K^{\prime} T^{\prime}$, so our answer is $2 \sqrt{5}$.