Limit of complex function is the same regardless of path taken.(Proof)

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(1) Suppose $f:\mathbb{C}\rightarrow$U and $\lim_{z \to z_0} f(z) = l$. I would like to show that irrespective of the path taken by the variable $z \to z_0$ the limit remains the same.

I have proven that if $\lim_{z \to z_0} f(z) = l$ iff $\lim f(z_n) = l$ $\forall (z_n)_n$ in $\mathbb{C}$ s.t $\lim_{n \to \infty} z_n=z_0$.

Is that sufficient to proof (1)?

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Let $\epsilon > 0$ and let $\delta>0$ be such that $|f(z)-L| < \epsilon$ if $|z-z_0| < \delta$.

If $c:[0,1]\to \Bbb C$ is a path that is continuous at $1$ and $c(t)\to z_0$ as $t\to 1^-$, then for all $t$ sufficiently close to $1$ we have $|z_0 - c(t)| < \delta$, so $|f(c(t)) - L| < \epsilon$. Hence the limit is independent of which path you take to approach $z_0$.