I was trying to derive surface area of a sphere myself. I started with a hemi-sphere, sliced it into infinite triangles and then added the area of all the triangles and finally doubled it to get area of a sphere. I'm getting π²R² as final result. I could not understand what's wrong with my logic.
The problem is that your pieces are not triangles and don't "approach triangles in the limit" in the necessary way. For example, the proportion of shaded area above height $\frac h2$ in your figure doesn't approach $\frac14$ of the total shaded area as you cut the hemisphere into smaller pieces; it remains somewhat less. Your approach actually computes the surface area of a cone.
Another way to see that your pieces are not triangles: the angle between the base and the sides is always 90°. This means the sides are curved, and each piece is fatter at the bottom than a triangle. Cutting it into thinner pieces doesn't change this; the pieces always have more area than the corresponding triangle, by a fixed ratio that does not approach 1.
As noted in the comments and Karl's (+1) answer, the issue is in flattening the "gores," which does not preserve area.
One correct reckoning is instead to use a quasi-triangle with two right angles at the base, angle $d\theta$ at the apex, and width $r\sin\varphi\, r\, d\theta$ at height $\varphi$ (below, left), as noted by md2perpe and bubba in the comments. The area of the flat region is correct, $$ \int_{0}^{\pi/2} r\sin\varphi\, d\varphi\, (r\, d\theta) = r^{2}\, d\theta, $$ though this region does not wrap without distortion (i.e., not isometrically) onto a gore, but only in an area-preserving way.
A related correct accounting (below, right, which shows a unit sphere; for radius $r$, multiply all lengths by $r$) slices the quasi-triangle by latitudes. The portion of a sphere between longitude $\theta$ and $\theta + d\theta$, and between colatitude $\varphi$ and $\varphi + d\varphi$, is an infinitesimal trapezoid. Project axially (away from the axis through the poles) onto a circumscribed cylinder. Up to an error infinitesimal compared to $d\theta\, d\varphi$, the trapezoid is a rectangle with width $\sin\varphi\, d\theta$ and height (measured along the sphere) $d\varphi$. The image of this quadrilateral under axial projection is a curved (but developable) rectangle of width $d\theta$ and height (along the cylinder) $\sin\varphi\, d\varphi$. Consequently, axial projection from a sphere to a circumscribed cylinder is area-preserving. (!!) Particularly, the area of a sphere is the area of a circumscribed cylinder, $(2\pi r)(2r) = 4\pi r^{2}$.
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After suitable scaling, the sides of the pseudo-triangle shown in the answer from Andrew Hwang are the graphs of $\cos(x)$ and $-\cos(x)$ curves, over the range $0$ to $\tfrac12 \pi$. By integrating the cosine function over this range, it's easy to show that the pseudo-triangle has $4/\pi$ times the area of the corresponding triangle.
The OP got the answer $\pi^2 R^2$ by adding up the areas of the triangles. If you multiply this by the area ratio $4/\pi$, you get the correct answer: $\pi^2 R^2 \times 4/\pi = 4\pi R^2$.
