This step is part of a larger proof. My linear algebra professor wrote that for $L:V \to W$,
$$ \dim(W)=\dim(\mathrm{range}(L)) \qquad\implies\qquad W = \mathrm{range}(L) $$
for finite dimensional $W$. If it is not clear $L$ is a linear map between two vector spaces $V$, $W$.
I am not sure why this follows. For example, two different lines in a plane have the same dimension of $1$. So clearly for general $A, B$ $$\dim(A) = \dim(B) \nRightarrow A = B$$
Let $v_1,\dots,v_m$ be a basis of range $T$. If $\dim \text{range }T=\dim W$, this means that a basis of range $T$ is a basis of $W$ as range $T\subset W$ and the basis of range $T$ is a linearly independent list of length $\dim W$ of vectors from $W$ and thus is a basis of $W$. By definition, a basis $w_1,\dots,w_n$ of a vector space $W$ is a linearly independent list of vectors in that vector space such that $\text{span}(w_1,\dots,w_n)=W$. Thus, if a basis of range $T$ is a basis of $W$, then we see that $\text{span}(v_1,\dots,v_m)=\text{range }T=W$. Thus, range $T=W$.
Your second example fails because a basis of $A$ is not a basis of $B$ as $b\not\subset A$.
To properly prove that if $$\dim \text{range }T=\dim W\implies \text{range }T=W$$
first prove that a basis of range $T$ is a basis of $W$(you can use the theorem that any linearly independent list of length $\dim W$ consisting of vectors from $W$ is a basis of $W$). Then just use the definition of a basis of a vector space to conclude that range $T=W$.
Let $R_L=\{L(x)|\ x\in V\}$ be range of $L$. We show $R_L$ is subspace of $W$. $(1)$ Since $L$ is a linear map, we have $L(0_V)=0_W$. Thus $0_W\in R_L$. $(2)$ let $y_1,y_2\in R_L$. Then $\exists x_1,x_2\in X$ such that $L(x_1)=y_1$ and $L(x_2)=y_2$. Since $L$ is linear map, we have $y_1+y_2 =L(x_1)+L(x_2)=L(x_1+x_2)$. Thus $y_1+y_2\in R_L$. $(3)$ let $c\in F$ and $y\in R_L$. Then $\exists x\in X$ such that $L(x)=y$. Since $L$ is linear map, we have $c\cdot y=c\cdot L(x)=L(c\cdot x)$. Thus $c\cdot y\in R_L$. Hence $R_L$ is subspace of $W$.
It’s a standard result, if $R_L$ is subspace of $W$ and $\text{dim}(R_L)=\text{dim}(W)\in \Bbb{N}$, then $R_L=W$. We use this result countless times, for instance see.
