Assume you have a map $\phi : V \rightarrow \mathbb{C}$, where $V$ is a complex vector space.
Now, if we have $\phi(\lambda x) = |\lambda | \phi(x)$ and $\phi(x+y)^2+ \phi(x-y)^2 = 2\phi(x)^2 +2 \phi(y)^2$. Is this sufficient to conclude that $\phi$ is a continuous map? Apparently it would be sufficient if we were able to conclude the triangle inequality from this I guess, but I don't see how this is possible.
Here is a rough draft of what I have tried so far and where the holes are. I will add to it as I find things relevant.
Let $b(x,y)=\phi(x+y)^2-\phi(x)^2-\phi(y)^2$, and note that $b(x,x)=2\phi(x)^2$ (the idea is that $b$ is twice the real part of a complex bilinear form. The imaginary part, and hence the form can be recovered from this if $b$ is bilinear).
Then $$b(x+y,z)=\phi(x+y+z)^2-\phi(x+y)^2-\phi(z)^2\\ =\left[-\phi(x-y)^2+2\phi\left(x+\frac{z}{2}\right)^2+2\phi\left(y+\frac{z}{2}\right)^2\right]-\phi(x+y)^2-4\phi\left(\frac{z}{2}\right)^2\\ =2\left[\phi\left(x+\frac{z}{2}\right)^2-\phi(x)^2-\phi\left(\frac{z}{2}\right)^2\right]+2\left[\phi\left(y+\frac{z}{2}\right)^2-\phi(y)^2-\phi\left(\frac{z}{2}\right)^2\right]\\ =2b\left(x,\frac{z}{2}\right)+2b\left(y,\frac{z}{2}\right)$$
If we let $y=0$, then we get $$b(x,z)=2b\left(x,\frac{z}{2}\right).$$ Plugging this into the last line of the derivation gives
$$b(x+y,z)=b(x,z)+b(y,z).$$ It is obvious that $b$ is symmetric, so $b$ is additive in each coordinate. Let $1>q>0$ have a finite dyadic expansion $q=\sum_{j=1}^M\frac{d_j}{2^j}$. Then we have $b(x,qz)=qb(x,z)$ by a finite number of applications of the rule $\frac{b(x,z)}{2}=b\left(x,\frac{z}{2}\right)$ and additivity. Note that additivity in each coordinate implies that $b(x,-y)=-b(x,y)$.
Let $1>r>0$ be real. $$rb(x,y)-b(x,ry)=(r-q)b(x,y)-b(x,(r-q)y)$$ We want to make the left hand side equal $0$. We can easily make $\vert r-q\vert$ small by choosing good dyadic approximations $q$ of $r$.
If we can prove that the second coordinate of $b$ is continuous at zero, then $b$ will be $\mathbb{R}$-linear in its second coordinate by $b(x,ay)=b(x,(a-\lfloor a\rfloor)y)+b(x,\lfloor a\rfloor y)=(a-\lfloor a\rfloor)b(x,y)+\lfloor a\rfloor b(x,y)=ab(x,y)$ for all real numbers $a$.
It is here that I assume that $V$ is finite dimensional: Linearity in the second coordinate implies that $b$ is bilinear (because $b$ is symmetric), this implies that it is continuous on $V\times V$, in particular on the diagonal. This implies that $\phi$ is continuous.