Is there a very small gap or no gap in this proof? ("Linear Algebra Done Right 3rd Edition" by Sheldon Axler.)

I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.

Let $$V$$ be a vector space.
Let $$V'$$ be the dual space of $$V$$.
Let $$W$$ be a vector space.
Let $$W'$$ be the dual space of $$W$$.

Definition:
If $$T\in\mathcal{L}(V,W)$$, then the dual map of $$T$$ is the linear map $$T'\in\mathcal{L}(W',V')$$ defined by $$T'(\varphi)=\varphi\circ T$$ for $$\varphi\in W'$$.

Definition:
For $$U\subset V$$. The annihilator of $$U$$, denoted $$U^0$$, is defined by $$U^0=\{\varphi\in V' : \varphi(u)=0\text{ for all }u\in U\}.$$

The next result is from "Linear Algebra Done Right 3rd Edition". (on p.107)

3.108 $$T$$ surjective is equivalent to $$T'$$ injective
Suppose $$V$$ and $$W$$ are finite-dimensional and $$T\in\mathcal{L}(V,W)$$. Then $$T$$ is surjective if and only if $$T'$$ is injective.

Proof The map $$T\in\mathcal{L}(V,W)$$ is surjective if and only if $$\operatorname{range} T=W$$, which happens if and only if $$(\operatorname{range} T)^0=\{0\}$$, which happens if and only if $$\operatorname{null} T'=\{0\}$$ [by 3.107(a)], which happens if and only if $$T'$$ is injective.

3.107(a) is the following equation:

If $$T\in\mathcal{L}(V,W)$$, $$\operatorname{null} T'=(\operatorname{range} T)^0.$$

The author is very kind to the readers and, usually, there is no gaps in his proofs in this book.
But I felt a small gap in the proof of 3.108 above.
I felt the following fact is not so obvious.

Fact 1:
If $$(\operatorname{range} T)^0=\{0\}$$, then $$\operatorname{range} T=W$$.

Proof:
Assume that $$(\operatorname{range} T)^0=\{0\}$$ but $$\operatorname{range} T\neq W$$.
Then, $$\dim \operatorname{range} T < \dim W$$.
Let $$v_1,\dots,v_k$$ be a basis of $$\operatorname{range} T$$.
Let $$v_1,\dots,v_k,\dots,v_l$$ be a basis of $$W$$ ($$k).
Let $$\varphi\in W'$$ be a linear functional such that
$$\varphi(v_i)=0$$ for all $$i\in\{1,\dots,l-1\}$$ and
$$\varphi(v_l)=1$$.
Then, $$\varphi\neq 0$$ and $$\varphi(v)=0$$ for any $$v\in\operatorname{range} T$$.
So, $$0\neq\varphi\in(\operatorname{range} T)^0$$.

Does the above fact immediately follow from some famous result?

By the way, the author commented about 3.107(a) as follows:

The proof of part (a) of the result below does not use the hypothesis that $$V$$ and $$W$$ are finite-dimensional.

And in 3.108, the author assumed that $$V$$ and $$W$$ are finite-dimensional.
So, we need to use the assumption that $$V$$ and $$W$$ are finite-dimensional in the proof of 3.108.

I guess we need to use the assumption that $$W$$ is finite-dimensional to prove Fact 1 above.

I guess we don't need the assumption that $$V$$ is finite-dimensional to prove 3.108.

3.106 Suppose $$V$$ is finite-dimensional and $$U$$ is a subspace of $$V$$. Then $$\dim U+\dim U^0=\dim V.$$