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I know decidable languages are closed under concatenation and union. But i am having a hard time finding an counter argument (i think its wrong).

Can you help me find the right direction to proof/disproof this lemma ?

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Unfortunately, it isn't true. Let $L$ be the language consisting of the symbols $a$ and $b$ together with an undecidable set of strings in that alphabet. In this case, $L^*$ has all strings, and so it is decidable, but $L$ is not.

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    What do you mean by 'Unfortunately' ? – 2017-01-29
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    I had just meant that it would be unfortunate, if one had been hoping to prove that it was true. I was reacting to the title of your post, but I see that in the body of the post, you did say that you thought it was wrong. So we're on the same page. – 2017-01-29
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    Thank you for taking time and answering my question. – 2017-01-29
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    Another traditional example: Every language may be encoded over a unary alphabet (e.g., see words over $\{a, b\}$ as binary numbers, and encode a number $n$ by $a^n$). But the star of any unary language is regular. – 2017-01-29
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    @MichaëlCadilhac why is the star of any unary language is regular? – 2017-02-07
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    @dave see http://cs.stackexchange.com/questions/21765/kleene-star-of-an-infinite-unary-language-always-yields-a-regular-language/21775 – 2017-02-07
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    @dave It is not too difficult to see that $A$ is any set of words in a unary alphabet, then $A^*$ is eventually periodic, and all such languages are regular. – 2017-02-07