Closure Properties between different Language Classes
Yours is a correct intuition: $L_1 \cap L_2$ is indeed recognisable.

However, you can't say that such language is not decidable only because you failed to prove it decidable. In theory, there could be a more complex argument that you were not able to find, which could establish that $L_1 \cap L_2$ is actually decidable.

To prove it, you need a counterexample. It is usually helpful to try corner cases, first.

For instance if $L_1 = \Sigma^*$, which is regular, we get $L_1 \cap L_2 = L_2$ which is decidable only if $L_2$ is such. Hence, since recognizable but non decidable languages exist, we get a counterexample.

Further, note that if we took $L_1 = \emptyset$, which is regular, then $L_1 \cap L_2 = \emptyset$ which is decidable, whatever $L_2$ is.

Hence, from the given hypotheses, we can not conclude that $L_1 \cap L_2$ is decidable, nor that it is not. This is because we found counterexamples to both conjectures.