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Show $(ev_alpha)_{alpha∈mathbb{R}}$ is linearly independent

$(ev_alpha): mathbb{R}[X]rightarrowmathbb{R}, fmapsto f(alpha)$

So I need to show, maybe with a counter-example, that $lambda=0$ in $sumlambda ev_alpha=0$

How can I do it?

An uncountable set is linearly independent if and only if every finite subset is linearly independent.

So suppose $U subset mathbb{R}$ is finite and there exists $lambda_alpha$ for each $alpha in U$ such that $ displaystyle sum_{alpha in U} lambda_alpha ev_alpha = 0$.

Then by definition of $ev_alpha$ we have $displaystyle sum_{alpha in U} lambda_alpha f(alpha) = 0$ for any $f in mathbb{R}[X]$.

Choose some $alpha' in U$ and define $displaystyle f_{alpha'}(x) =prod_{alpha in U - { alpha' }}(x-alpha)$.

Then by construction we have $f_{alpha'}(alpha') neq 0$, but $f_{alpha'}(alpha) = 0$ for all $alpha in U - {alpha'}$.

Then $displaystyle 0= sum_{alpha in U} lambda_alpha f_{alpha'}(alpha) = lambda_{alpha'}f_{alpha'}(alpha')$ and since $f_{alpha'}(alpha')neq 0$ we have $lambda_{alpha'}=0$.

Since our choice of $alpha'$ was arbitrary, we have $lambda_alpha=0$ for all $alpha in U$.

Thus the finite subset $(ev_alpha)_{alpha in U}$ is linearly independent.

And since our choice of $U$ was arbitrary, every finite subset is linearly independent.

Therefore $(ev_alpha)_{alpha in mathbb{R}}$ is linearly independent.