What is wrong in the proof of max value of $|iz+z_0|$ 
If $|z-i|\leq 2,z_0 = 5+3i$ then find max value of $|iz+z_0|.$

Now ,
$|iz+z_0| \leq |iz|+|z_0| \cdots \text{(By Triangle Inequality.)}$
Now consider the diagram in Argand Plane -

Observe that $|z| \leq 3.$
$\Rightarrow |iz+z_0| \leq \boxed{3 + \sqrt{34}}$
But the Answer is $\boxed{7}.$ What is wrong in this proof $?$

Where you were incorrect is not accounting for the change of $z \to iz$. While its true that $|z| = |iz|$, this does not always apply when adding complex numbers, $|z + z_0| ≠ |iz + z_0|$.
In your method, when you substituted $|iz| = 3$, you assimilated that similar substitution also applies to $|iz + z_0|$ towards, as you would for $|z + z_0|$.
If $z = x + yi$, then $|z + z_0| = |x + 5 + (y + 3)i|$; its easy to see that you would choose a set of possible positive ${x, y}$ within the bound $|z - i| \le 2$, while obtaining the longest length ($|z + z_0|_{max}$).
However, $|iz + z_0| = |xi - y + 3i + 5|$. Notice the $y$ is now negated. This means that using positive $y$ values, will reduce the coeff of the complex number $iz + z_0$ (as $x_{z0} > 0$ and $y_{z0} > 0$) and decrease $|iz + z_0|$, instead of obtaining the maximum.
Instead, you need a set of ${x, -y}$ to obtain $|iz + z_0|_{max}$. [$x$ is unchanged as rotation is only by $\frac{\pi}{2}$, $z\to iz$].
Graphically, your solution lies in 4 quadrant (sector) of the $|z - i| = 2$:

While I did not yet solve it algebraically, I tried trial and error and reached the approximation of $\approx 7$


Finding an exact answer is left as an exercise.

UPDATE
While the the triangle inequality is not used in the method below, consider if it might suit the question better.
From above, we know the solution lies on the arc of the curve, $y = -\sqrt{4 - x^2} + 1, x \in [0, \sqrt{3}]$
Hence, substituting this for $x, y$ in the complex number $iz$, we get:
$$|iz + z_0|$$
$$\to |ix - y + 5 + 3i|$$
$$\to |ix - \big[-\sqrt{4 - x^2} + 1 \big] + 5 + 3i|$$
$$\to |(x + 3)i + (\sqrt{4 - x^2} + 4)|$$
$$\to \sqrt{(x + 3)^2 + \big[ \sqrt{4 - x^2} + 4 \big]^2}$$
$$\to \sqrt{8\sqrt{4 - x^2} + 6x + 29}$$
In order to find the maximum length, we need to find where the length function above reaches a maximum.
Hence,
$$\frac{d}{dx}\sqrt{8\sqrt{4 - x^2} + 6x + 29} = 0$$
$$\frac{d}{dx}\big[8\sqrt{4 - x^2} + 6x + 29\big] \cdot \frac{1}{\sqrt{8\sqrt{4 - x^2} + 6x + 29}} = 0$$
We have numerator = 0,
$$\to 3\sqrt{4 - x^2} + 4x = 0$$
$$\therefore x = \frac{6}{5}, x \in [0, \sqrt{3}]$$
Finally,
$$\to |(\frac{6}{5} + 3)i + (\sqrt{4 - \big(\frac{-6}{5}\big)^2} + 4)|$$
$$ = 7 \text{, as required}$$
