If every sequence $(x_{n}) \subset X$ and $(\lambda_{n}) \subset \mathbb{R}$ we have $\lim \lambda_{n}x_{n} = 0$, then $X$ is bounded

Your part (a) is fine, assuming that you know translation is continuous.

Your part (b) is on the right track, but it assumes that $V$ is the continuum, whereas it should be a normed vector space. You have the right idea of separating two sets by a very small set, but you shouldn't rely on coordinates (since you don't have any coordinates...). Use the norm instead.

Your part (c) has a similar issue: unless you're in a finite dimensional space, there is neither reason nor justification for writing the sequence in coordinates. Again, you need to be using the norm. Notice that

$$\|\lambda_n x_n\| = |\lambda_n| \cdot \|x_n\| \le |\lambda_n| \sup_{x \in X} \|x\| \to 0.$$

The converse is in fact true, and can be handled similarly; choose an unbounded sequence $\{x_n\}$ and craft $\lambda_n$ depending on $x_n$ so that $\|\lambda_n x_n\| \not\to 0$. Constant norm works.

For the converse: suppose $X$ is not bounded. Then it contains some sequence $(x_n)$ such that $(\|x_n\|)\to\infty$ (take $x_n$ to be an example given by the negation of the definition of "bounded" with the constant set to $n$). Take $\lambda_n = \frac{1}{\|x_n\|}$. Then $(\lambda_n)\to 0$, but $(\lambda_nx_n)\not\to 0$, since $(\|\lambda_nx_n\|) = (1) \not\to 0$. Thus, taking the contrapositive, we have the converse (NB: you've got a bit missing out of the contrapositive you've given in the title: you've dropped the $\lim\lambda_n = 0$ bit).