I am trying to understand how to show two vector spaces are isomorphic. You do this by showing there is an isomorphism that can be mapped between the two spaces.
What I don't understand is my lecturer's way of showing that the isomorphism is one-to-one?
It seems to me that he only showed that f is one-to one for the case of $w=0$. This, doesn't generalize to other points does it?
In general, showing a map $f:A \to B$ is one-to-one involves showing that for any $x,y \in A$ such that $x \ne y$, we have $f(x) \ne f(y)$. However, an isomorphism between vector spaces is by definition a linear map, and your lecturer should have shown (maybe much earlier) that a linear map is one-to-one if and only if the kernel of $f$ is $\{0\}$, i.e., $f$ maps nothing to zero except zero. Indeed, this follows from the fact that $f(x)-f(y)=f(x-y)$ if $f$ is linear.
Proof of last claim:
Let $f$ be linear and one-to-one. Suppose for sake of contradiction that the kernel of $f$ is nontrivial, i.e., there exists some $x \ne 0$ such that $f(x)=0$. Then for any $y$ we have $f(y+x)=f(y)+f(x)=f(y)$ but $y+x \ne y$, contradicting the fact that $f$ is one-to-one.
For the converse, let $f$ be linear and have kernel $\{0\}$. Let $a \ne b$. Then $f(b)-f(a) = f(b-a) \ne 0$ because $b-a \ne 0$; thus $f$ is one-to-one.
Let $V$ and $W$ be vector spaces and let $T:V\to W$ be a linear transformation. In general, $T$ is one-to-one if and only if $$\ker{T} = \{v\in V\ |\ Tv=0_W\} = \{0_V\},$$ where $0_V$ is the zero vector in $V$. The fact that you may be looking at is that $T$ is one-to-one if and only if $w\in\ker{T}$ implies $w=0_V$.