How to determine the sign of passive elements in mesh analysis?
Your professor is correct.

Current down through \$\ C_1 ,\: R_2\$ is: \$\ (gV_C-I)\$, hence \$V_0\$ for this branch is:

$$V_0=(gV_C-I)(10+\frac{10}{s})$$

Voltage across \$R_3\$ is \$\ 5I\$, hence \$V_0\$ for the right branch is:

$$V_0=5I+V_S$$

Equating these:

$$(gV_C-I)(10+\frac{10}{s})-5I-V_S=0$$

Multiply through by -1:

$$(I-gV_C)(10+\frac{10}{s})+5I+V_S=0$$