It's positive-valued and absolutely homogeneous, but I can't prove it satisfies the triangle inequality. I tried many "counter examples", but all of them satisify the inequality.
If we take T and K matrices, we get to this expression $\sqrt{||T^TT + K^TT + T^TK + K^TK||_2}$. And I don't know how to carry on from here
In general, if you have $$|a+b|\le |a|+|b|,$$ then you also have $$\sqrt{|a+b|}\le\sqrt{|a|+|b|}\le\sqrt{|a|}+\sqrt{|b|}.$$ This means that for any norm $\|\cdot\|,$ its square root is also a norm, regardless of whether you are on a matrix space etc.
EDIT: Actually, what I wrote above is non-sense as shown in the comments. The right way to proceed is to either know/derive the following fact: $$\|A^TA\|_2 = \|AA^T\|_2 = \sigma(AA^T) = \sigma(A)^2=\|A\|_2^2$$ and then it's done. I am assuming you are using the usual induced $2$-norm.