Let $V$ and $W$ be vector spaces and $T:V\rightarrow W$ be linear.
(a) Prove that $T$ is one-to-one iff $T$ carries linearly independent subsets of $V$ onto linearly independent subsets of $W$.
(b) Suppose that $T$ is one-to-one and that $S$ is a subset of $V$. Prove that $S$ is linear independent iff $T(S)$ is linear independent.
(c) Suppose $\mathcal{B} = \{v_{1},v_{2},\ldots,v_{n}\}$ is a basis for $V$ and $T$ is one-to-one and onto. Prove that $T(\mathcal{B}) = \{T(v_{1}),T(v_{2}),\ldots,T(v_{n})\}$ is a basis for $W$.
MY ATTEMPT (EDIT)
(a) We shall prove that implication ($\Rightarrow$) first.
Suppose that $S = \{s_{1},s_{2},\ldots,s_{m}\}\subset V$ is linear independent and $T$ is injective. Then we have that \begin{align*} & a_{1}T(s_{1}) + a_{2}T(s_{2}) + \ldots + a_{m}T(s_{m}) = 0 \Longrightarrow\\\\ & T(a_{1}s_{1} + a_{2}s_{2} + \ldots + a_{m}s_{m}) = 0 \Longrightarrow\\\\ & a_{1}s_{1} + a_{2}s_{2} + \ldots + a_{m}s_{m} = 0 \Longrightarrow\\\\ & a_{1} = a_{2} = \ldots = a_{m} = 0 \end{align*}
Thus $\{T(s_{1}),T(s_{2}),\ldots,T(s_{m})\}$ is linear independent, just as it has been claimed.
Let us prove that $(\Leftarrow)$ holds too.
We must prove that $T(v) = 0$ implies that $v = 0$. Indeed, according to the given assumption, if $\mathcal{B}_{V} = \{v_{1},v_{2},\ldots,v_{n}\}$ is a basis for $V$, then $T(\mathcal{B}_{V}) = \{T(v_{1}),T(v_{2}),\ldots,T(v_{n})\}$ is linear independent. Thence we get that \begin{align*} & T(v) = T(a_{1}v_{1} + a_{2}v_{2} + \ldots + a_{n}v_{n}) = 0 \Longrightarrow\\\\ & a_{1}T(v_{1}) + a_{2}T(v_{2}) + \ldots + a_{n}T(v_{n}) = 0 \Longrightarrow\\\\ & a_{1} = a_{2} = \ldots = a_{n} = 0 \Longrightarrow v = 0 \end{align*} and we are done.
(b) Let us prove $(\Rightarrow)$ first.
If $S$ is LI, then $T(S)$ is LI, according to (a).
Let us now prove the implication $(\Leftarrow)$.
If $S = \{s_{1},s_{2},\ldots,s_{n}\}$, then $T(S) = \{T(s_{1}),T(s_{2}),\ldots,T(s_{n})\}$. If $T(S)$ is LI, then one has that \begin{align*} & a_{1}s_{1} + a_{2}s_{2} + \ldots + a_{n}s_{n} = 0 \Longrightarrow\\\\ & T(a_{1}s_{1} + a_{2}s_{2} + \ldots + a_{n}s{n}) = 0 \Longrightarrow\\\\ & a_{1}T(s_{1}) + a_{2}T(s_{2}) + \ldots + a_{n}T(s_{n}) = 0 \Longrightarrow\\\\ & a_{1} = a_{2} = \ldots = a_{n} = 0 \end{align*} Thus $S$ is linear independent, just as desired.
(c) If $T$ is one-to-one, then $T(\mathcal{B}) = \{T(v_{1}),T(v_{2}),\ldots,T(v_{n})\}$ is linear independent. This is because $\mathcal{B}$ is LI as well.
Since $T$ is onto, we have that $T(V) = \text{span}(T(\mathcal{B})) = W$. Consequently, $T(\mathcal{B})$ is a basis for $W$, and we are done.
Could someone double-check my arguments?
Your revised attempt to show that if $T$ sends linearly independent sets to linearly independent sets is correct as far as it goes. It suffers form an unwarranted assumption that your space is finite dimensional (though you are probably working just with finite dimensional spaces). This can easily be fixed by taking an arbitrary basis and then just taking finitely many basis vectors needed to express $v$.
But there is a much simpler way: it is enough to show that if $v\neq 0$ then $T(v)\neq 0$. If $v\neq 0$, then $\{v\}$ is linearly independent, hence $\{T(v)\}$ is linearly independent; hence $T(v)\neq 0$.
Alternatively, using contrapositives, if $T(v)=0$ then $\{T(v)\}$ is linearly dependent, which means that $\{v\}$ is linearly dependent, which means that $v=0$.