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It is given that $\mathcal{B}$ is a base for the metric space $(X,d)$. Now our claim is that $\mathcal{A}=\{G\in\mathcal{B}| a\in G\}$ forms a neighborhood base of $a$, where $a\in X$.

The attempted justification of the claim:

$\mathcal{A}\neq \emptyset$ is an easy observation. Let $U$ be a neighborhood of $a$. Then $U=\bigcup\{G_\lambda|\lambda\in\Lambda\}$ for some subcollection $\{G_\lambda|\lambda\in\Lambda\}$ in $\mathcal{A}$ $(\subseteq\mathcal{B})$. (Here I am taking the neighborhoods of a point to be open sets containing that point). Thus $G_\lambda{_0}\subseteq U$ for some $\lambda_0\in\Lambda$. Hence $\mathcal{A}$ is a neighborhood base of $a$.

Is the above "justification" a justification?

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    If you change $G_{\lambda0}\subset U$ to $a\in G_{\lambda0}$, then it would be correct. – 2017-02-15
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    By default $a\in G_\lambda{_0}$ is true, isn't it? What I've tried to prove is that for any neighborhood of $a$, there is a neighborhood of $a$ in $\mathcal{A}$ which is contained in the aforementioned neighborhood of $a$. Am I wrong? – 2017-02-15

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It's wrong. When you say $\lbrace G_{\lambda}|\lambda\in\Lambda\rbrace$ in $\mathcal{A}$ you're using the claim. So you need to say it's in $\mathcal{B}$ and then $a\in U\implies\exists \lambda_{0}:a\in G_{\lambda_{0}}$ and then $G_{\lambda_0}\in\mathcal{A}$