Your proof looks correct, and is well presented. My only (rather minor) suggestion is that having a diagram would have made it a bit easier to follow.
FYI, here's an alternate proof method. In $\triangle ABC$, let the side lengths opposite $A$, $B$ and $C$ be $a$, $b$ and $c$, respectively. The law of cosines states that
$$a^2 = b^2 + c^2 - 2bc\cos(A) \tag{1}\label{eq1A}$$
Also, in $\triangle AB'C'$, have the side lengths opposite $A$, $B'$ and $C'$ be $a'$, $b'$ and $c'$, respectively. This then gives
$$(a')^2 = (b')^2 + (c')^2 - 2(b')(c')\cos(A) \tag{2}\label{eq2A}$$
Note $b' \gt b \;\;\to\;\; (b')^2 \gt b^2$ and $c' \gt c \;\;\to\;\; (c')^2 \gt c^2$. Also, since $\measuredangle A$ is a right or obtuse angle, then $\cos(A) \le 0 \;\;\to\;\; -\cos(A) \ge 0$, so $- 2(b')(c')\cos(A) \ge - 2bc\cos(A)$. Thus, the RHS of \eqref{eq2A} is greater than the RHS of \eqref{eq1A}, which means $(a')^2 \gt a^2 \;\;\to\;\; a' \gt a$.
Here's a second alternative proof method. With $\triangle ABC$, using the same side lengths as with \eqref{eq1A}, the law of sines says that
$$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \tag{3}\label{eq3A}$$
and, similarly with $\triangle AB'C'$, we have
$$\frac{a'}{\sin(A)} = \frac{b'}{\sin(B')} = \frac{c'}{\sin(C')} \tag{4}\label{eq4A}$$
Note the angles $B'$ and $C'$, compared to $B$ and $C$, respectively, have either not changed, or one of them has decreased and the other increased. WLOG, have $\measuredangle B' \le \measuredangle B$. Since $\sin(x)$ is a strictly increasing function for $0^{\circ} \lt x \lt 90^{\circ}$, then $\sin(B') \le \sin(B) \;\to\; \frac{1}{\sin(B')} \ge \frac{1}{\sin(B)}$. Since $b' \gt b'$, then comparing \eqref{eq4A} to \eqref{eq3A}, we get $\frac{b'}{\sin(B')} \gt \frac{b}{\sin(B)} \;\to\; \frac{a'}{\sin(A)} \gt \frac{a}{\sin(A)} \;\to\; a' \gt a$.
Best Answer
This is not true. See the diagram below:
Comments on the Edited Question
As mentioned in comments under Pythagoras' answer, it is true that if $a\gt b$, then $A\gt B$. To show this, consider the incircle of our triangle:
Looking at the right triangles $\triangle OEC$ and $\triangle OEB$, we have $$ \begin{align} a &=CE+EB\tag{1a}\\ &=r\cot(C/2)+r\cot(B/2)\tag{1b} \end{align} $$ Looking at the right triangles $\triangle ODC$ and $\triangle ODA$, we have $$ \begin{align} b &=CD+DA\tag{2a}\\ &=r\cot(C/2)+r\cot(A/2)\tag{2b} \end{align} $$ Thus, if $a\gt b$, then $\cot(B/2)\gt\cot(A/2)$, which, since $\cot(\theta)$ is decreasing on $(0,\pi/2)$, says that $A\gt B$.
Therefore, since the sum of the three angles must be $180^{\large\circ}$, $B$ must be less than $90^{\large\circ}$; otherwise, the sum of $A$ and $B$ would be greater than $180^{\large\circ}$.