Probability of a person correctly guessing at least one number out of the two number another person chooses 
Person A randomly chooses a number from 1 to 5 (inclusive) twice, so A ends up with 2 numbers chosen (can be the same number). Person B also makes a random choice from that list (only 1 number). What's the probability that B's choice match at least one of A's choices?

My interpretation of the question is: what's the probability that a randomly chosen number (call it $c$) between 1 and 5 is in the random set $[a, b],~ a $ and $b $are between 1 and 5.
My attempt of solving this question is the following:
There are 2 cases: 1.$ a$ and $b$ are different numbers; 2. $a$ and $b$ are the same.\
case 1:
if $a = b$, $\mathbb{P}[c \in \{a, b\}] = \mathbb{P}[c = a] = \frac{1}{5} $
case 2:
if $a \not= b$, $\mathbb{P}[c \in \{a, b\}] = \mathbb{P}[c = a] + \mathbb{P}[c = b]= \frac{2}{5} $
The probability of case 1 occurring is $\frac{1}{5}$. The calculation is similar to case 1. And the probability of case 2 occurring is $\frac{4}{5}$ since its the complement of case 1. Then, the answer is $$\mathbb{P}[\text{case 1}] \cdot\frac{1}{5} + \mathbb{P}[\text{case 2}]\cdot\frac{2}{5} = \frac{1}{5} \cdot \frac{1}{5} + \frac{4}{5} \cdot\frac{2}{5} = \frac{3}{5}$$
However, the correct answer is $\frac{9}{25}$. This is confirmed by a simulation I ran.
What did I do wrong?

Your reasoning is correct up to the very last line:
$$\mathbb{P}[\text{case 1}] \cdot\frac{1}{5} + \mathbb{P}[\text{case 2}]\cdot\frac{2}{5} = \frac{1}{5} \cdot \frac{1}{5} + \frac{4}{5} \cdot\frac{2}{5}$$
But this is not equal to $\frac{3}{5}$.
Instead:
$$\frac{1}{5} \cdot \frac{1}{5} + \frac{4}{5} \cdot\frac{2}{5} = \frac{1 \cdot 1}{25} + \frac{4 \cdot 2}{25} = \frac{9}{25}$$
