Consider a game $G=(S_1,...,S_n;u_1,...,u_n)$
Suppose $a$ and $b$ are strategies for a player, and let $\lambda \in [0,1]$. Show that $\lambda a+(1-\lambda)b$ is also a strategy. Now I want to first get an intuitive feel for what I am proving. If $\lambda$=0 then $\lambda a+(1-\lambda)b=b$ and if $\lambda=1$ then $\lambda a+(1-\lambda)b=a$. Suppose $a=1$ and $b=1$, then any $\lambda$ value will give $\lambda a+(1-\lambda)b=1$, which makes sense intuitively.
Now $a$ and $b$ are just probability values, so it seems intuitive that a convex combination would also be a strategy. Nevertheless, I am struggling to prove it, rather I seem to go around in circles and come up with tautologies.
My understanding is that I am trying to prove that since $a,b$ lie between [0,1], and $\lambda$ also lies in this range , then this necessarily implies $\lambda a+(1-\lambda)b$ lies between $[0,1]$, otherwise it would not be strategy.
So let's for the time being assume that, for all agents $i$, there are only finitely many possible actions, i.e. that $|S_i|<\infty$. Then for each $i$, we can identify the set of all mixed strategies corresponding to the set of pure strategies $S_i$ with the $(|S_i|-1)$-dimensional simplex in Euclidean space.
Lets say $|S_i|= n+1$. Then a probability distribution on $S_i$ is simply an assignment of weakly positive numbers $p_1, \ldots, p_{n+1}$, one to each of the pure strategies respectively, such that: $$ \sum_{k=1}^{n+1} p_k = 1. $$ Now this constraint that the sum of all the probabilities is 1 takes away one degree of freedom. Let's say that we have an assignment of $p_1, \ldots, p_n$ such that $\sum_{k=1}^n p_k < 1$. Then to be a valid probability distribution it must be the case that $p_{n+1} = 1-\sum_{k=1}^n p_k$.
So take as an example a very simple case: say our game is matching pennies and hence my strategy set is $\{H, T\}$. Then I can pick any assignment $0 \le p_H\le 1$ for the probability a mixed strategy places on heads, but that necessarily uniquely determines a probability distribution over $\{H,T\}$ because $p_T$ must equal $1-p_H$. Hence I can identify the set of all mixed possible mixed strategies for this strategy space with the set of all real numbers between 0 and 1 inclusive: the possible reasonable values of $p_H$.
Now, say I have two distributions, $\alpha=(\alpha_1, \ldots, \alpha_{n+1})$ and $\beta=(\beta_1, \ldots, \beta_{n+1})$, over our strategy space's $n+1$ actions. Then because they are probability distributions: $$ \sum_{k=1}^{n+1} \alpha_k = \sum_{k=1}^{n+1} \beta_k = 1 $$ So to show convexity now, as you said, we want to show that an interior combination of our arbitrary probability distributions is also a valid probability distribution. Clearly if $\forall k$ we have $\alpha_k,\beta_k \ge 0$ so too is $\lambda \alpha_k + (1-\lambda)\beta_k$. So all that remains is to check the sum of the probabilities in $\gamma = \lambda \alpha + (1-\lambda)\beta$ is 1.
But: $$ \sum_{k=1}^{n+1} \gamma_k = \sum_{k=1}^{n+1}\big[ \lambda \alpha_k+ (1-\lambda)\beta_k \big] = \lambda \sum_{k=1}^{n+1} \alpha_k + (1-\lambda)\sum_{k=1}^{n+1} \beta_k = \lambda + (1-\lambda)=1 $$ where the second-to-last equality comes from $\alpha$ and $\beta$ being probability distributions!
Very usefully, this is even true when $|S_i|$ is infinite! But this requires somewhat of an ascention in technique/language to discuss. If you're interested I'd be happy to elaborate.